Bashar was practicing for the national programming contest. Because of sitting too much in front of the computer without doing physical movements and eating a lot Bashar became much fatter. Bashar is going to quit programming after the national contest and he is going to become an actor (just like his father), so he should lose weight.

In order to lose weight, Bashar is going to run for k kilometers. Bashar is going to run in a place that looks like a grid of n rows and m columns. In this grid there are two one-way roads of one-kilometer length between each pair of adjacent by side cells, one road is going from the first cell to the second one, and the other road is going from the second cell to the first one. So, there are exactly (4 n m - 2n - 2m) roads.

Let's take, for example, n = 3 and m = 4. In this case, there are 34 roads. It is the picture of this case (arrows describe roads):

<image>

Bashar wants to run by these rules:

  * He starts at the top-left cell in the grid; 
  * In one move Bashar may go up (the symbol 'U'), down (the symbol 'D'), left (the symbol 'L') or right (the symbol 'R'). More formally, if he stands in the cell in the row i and in the column j, i.e. in the cell (i, j) he will move to: 
    * in the case 'U' to the cell (i-1, j); 
    * in the case 'D' to the cell (i+1, j); 
    * in the case 'L' to the cell (i, j-1); 
    * in the case 'R' to the cell (i, j+1); 
  * He wants to run exactly k kilometers, so he wants to make exactly k moves; 
  * Bashar can finish in any cell of the grid; 
  * He can't go out of the grid so at any moment of the time he should be on some cell; 
  * Bashar doesn't want to get bored while running so he must not visit the same road twice. But he can visit the same cell any number of times. 



Bashar asks you if it is possible to run by such rules. If it is possible, you should tell him how should he run.

You should give him a steps to do and since Bashar can't remember too many steps, a should not exceed 3000. In every step, you should give him an integer f and a string of moves s of length at most 4 which means that he should repeat the moves in the string s for f times. He will perform the steps in the order you print them.

For example, if the steps are 2 RUD, 3 UUL then the moves he is going to move are RUD + RUD + UUL + UUL + UUL = RUDRUDUULUULUUL.

Can you help him and give him a correct sequence of moves such that the total distance he will run is equal to k kilometers or say, that it is impossible?

Input

The only line contains three integers n, m and k (1 ≤ n, m ≤ 500, 1 ≤ k ≤ 10 ^{9}), which are the number of rows and the number of columns in the grid and the total distance Bashar wants to run.

Output

If there is no possible way to run k kilometers, print "NO" (without quotes), otherwise print "YES" (without quotes) in the first line.

If the answer is "YES", on the second line print an integer a (1 ≤ a ≤ 3000) — the number of steps, then print a lines describing the steps.

To describe a step, print an integer f (1 ≤ f ≤ 10^{9}) and a string of moves s of length at most 4. Every character in s should be 'U', 'D', 'L' or 'R'.

Bashar will start from the top-left cell. Make sure to move exactly k moves without visiting the same road twice and without going outside the grid. He can finish at any cell.

We can show that if it is possible to run exactly k kilometers, then it is possible to describe the path under such output constraints.

Examples

Input


3 3 4


Output


YES
2
2 R
2 L


Input


3 3 1000000000


Output


NO


Input


3 3 8


Output


YES
3
2 R
2 D
1 LLRR


Input


4 4 9


Output


YES
1
3 RLD


Input


3 4 16


Output


YES
8
3 R
3 L
1 D
3 R
1 D
1 U
3 L
1 D

Note

The moves Bashar is going to move in the first example are: "RRLL".

It is not possible to run 1000000000 kilometers in the second example because the total length of the roads is smaller and Bashar can't run the same road twice.

The moves Bashar is going to move in the third example are: "RRDDLLRR".

The moves Bashar is going to move in the fifth example are: "RRRLLLDRRRDULLLD". It is the picture of his run (the roads on this way are marked with red and numbered in the order of his running):

<image>
import java.io.*;
import java.math.*;
import java.util.*;

/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { 1, 0, -1, 0};
	private static int dy[] = { 0, -1, 0, 1};

	private static final long INF = (long) (1e15);
	private static final int INT_INF = Integer.MAX_VALUE;
	private static final long NEG_INF = Long.MIN_VALUE;
	private static final int NEG_INT_INF = Integer.MIN_VALUE;
	private static final double EPSILON = 1e-10;

	private static final int MAX = 1007;
	private static final long MOD = 1000000007;
	
	private static final int MAXN = 100007;
	private static final int MAXA = 10000009;
	private static final int MAXLOG = 22;

	public static void main(String[] args) {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		

/*



*/
		
		int n = in.nextInt();
		int m = in.nextInt();
		int k = in.nextInt();
		
		int count = 0;
		List<String> list = new ArrayList<String>();
//		StringBuilder sb = new StringBuilder();
		int val = 0;
		for(int i = 0; i < m - 1; i++) {
			count++;
			val++;
//			sb.append("R");
			if(count == k) {
				list.add(val + " R");
				val = 0;
				break;
			}
		}
		if(count != k) {
			if(val > 0) {
				list.add(val + " R");
			}
			val = 0;
			for(int i = 0; i < m - 1; i++) {
//				sb.append("L");
				count++;
				val++;
				if(count == k) {
					list.add(val + " L");
					val = 0;
					break;
				}
			}
			if(count != k && val > 0) {
				list.add(val + " L");
				val = 0;
			}
		}
		
		if(count != k) {
			val = 0;
			for(int i = 1; i < n; i++) {
				if(count == k) {
					break;
				}
				count++;
				
				if(m == 1) {
					val++;
					continue;
				}
				
				list.add(1 + " D");
				val = 0;
				if(count == k) {
					break;
				}
				
				for(int j = 0; j < m - 1; j++) {
					count++;
					val++;
					if(count == k) {
						list.add(val + " R");
						val = 0;
						break;
					}
				}
				if(count != k) {
					if(val > 0) {
						list.add(val + " R");
					}
					val = 0;
					for(int j = 0; j < m - 1; j++) {
						count++;
						val++;
						if(count == k) {
							break;
						}
						count++;
						val++;
						if(count == k) {
							break;
						}
						count++;
						val++;
						if(count == k) {
							break;
						}
					}
					if(val > 0 && val % 3 == 0) {
						list.add((val/3) + " " + "UDL");
					}
					else if(val > 3) {
						list.add((val/3) + " " + "UDL");
						val = val % 3;
						if(val == 1) {
							list.add(1 + " " + "U");
						}
						else if(val == 2) {
							list.add(1 + " " + "UD");
						}
					}
					else if(val > 0) {
						if(val == 1) {
							list.add(1 + " " + "U");
						}
						else if(val == 2) {
							list.add(1 + " " + "UD");
						}
					}
					
				}
			}
		}
		if(m == 1 && val > 0) {
			list.add(val + " D");
		}
		if(count != k) {
			val = 0;
			for(int i = 0; i < n - 1; i++) {
//				sb.append("U");
				count++;
				val++;
				if(count == k) {
					break;
				}
			}
			if(val > 0) {
				list.add(val + " U");
			}
			
		}
		
		
		if(count != k) {
			System.out.println("NO");
		}
		else {
			System.out.println("YES");
			System.out.println(list.size());
			for(String str : list) {
				System.out.println(str);
			}
		}
		
	
		out.flush();
		out.close();
		System.exit(0);
	}
	
	
	private static class Graph{
		private int node;
		private List<Edge>[] adj;
		
		Graph(int node){
			this.node = node;
			adj = new ArrayList[this.node];
			for(int i = 0; i < node; i++) {
				adj[i] = new ArrayList<Main.Edge>();
			}
		}
		
		void addEdge(int u, int v, long cost) {
			Edge e = new Edge(u, v, cost);
			Edge e1 = new Edge(v, u, cost);
			adj[u].add(e);
			adj[v].add(e1);
		}
		
		long[] calculateShortestPath(int source) {
			
			long dist[] = new long[node];
			PriorityQueue<Node> queue = new PriorityQueue<Main.Node>(new Comparator<Node>() {

				@Override
				public int compare(Node o1, Node o2) {
					int val = Long.compare(o1.distance, o2.distance);
					return val;
					
				}
			});
			
			Arrays.fill(dist, INT_INF);
			
			queue.add(new Node(source, 0));
			dist[source] = 0;
			
			while(!queue.isEmpty()) {
				Node nn = queue.remove();
				int u = nn.node;
				long udist = nn.distance;
				if(udist > dist[u]) {
					continue;
				}
				for(Edge e : adj[u]) {
						int v = e.v;
						long vdist = e.cost;
						long arrive = udist + vdist;
						if(arrive < dist[v]) {
							dist[v] = arrive;
							queue.add(new Node(v, dist[v]));
//							System.out.println(u + " " + v + " " + vdist + " " + dist[v]);
						}
					
				}
				
			}
			return dist;
			
		}
	}
	
	private static class Node{
		int operation;
		int node;
		long distance;
		Node(int operation, int node, long distance){
			this.operation = operation;
			this.node = node;
			this.distance = distance;
		}
		
		Node(int node, long distance){
			this.node = node;
			this.distance = distance;
		}
	}
	
	private static class Edge{
		int u;
		int v;
		long cost;
		
		Edge(int u, int v, long cost){
			this.u = u;
			this.v = v;
			this.cost = cost;
		}
		
		int getNeighbourIndex(int node) {
			if(this.u == node) {
				return v;
			}
			return u;
		}
	}
	
	private static boolean isPalindrome(String str) {
		StringBuilder sb = new StringBuilder();
		sb.append(str);
		String str1 = sb.reverse().toString();
		return str.equals(str1);
 	}

	private static String reverseString(String str) {
		StringBuilder sb = new StringBuilder();
		sb.append(str);
		return sb.reverse().toString();
	}
	
	
	private static double distance(Point p1, Point p2) {
		long divx = p2.x - p1.x;
		long divy = p2.y - p1.y;
		divx = divx * divx;
		divy = divy * divy;
		return Math.sqrt(divx + divy);
	}
	
	private static String getBinaryStr(int n, int j) {
		String str = Integer.toBinaryString(n);
		int k = str.length();
		for(int i = 1; i <= j - k; i++) {
			str = "0" + str;
		}
		
		return str;
	}

	// O(log(max(A,M)))
	static long modInverse(long a, long m) {
		extendedEuclid(a, m);
		return (x % m + m) % m;
	}

	static long x;
	static long y;
	static long gcdx;

	private static void extendedEuclid(long a, long b) {
		if (b == 0) {
			gcdx = a;
			x = 1;
			y = 0;
		} else {
			extendedEuclid(b, a % b);
			long temp = x;
			x = y;
			y = temp - ((a / b) * y);
		}
	}
	
	private static void generatePrime(int n) {
		 //O(NloglogN)
		boolean arr[] = new boolean[n + 5];
		Arrays.fill(arr, true);
		for(int i = 2; i * i <=n; i++){
			if(arr[i] == true){
				for(int j = i * i; j <= n; j+=i){
					arr[j] = false; 
				}
			}
		}
		int count = 0;
		int start = 0;
		for(int i = 2; i <= n; i++){
			if(arr[i] == true){
//				System.out.println(i + " ");
				count++;
			}
			if(count == (start * 100) + 1) {
//				System.out.println(i);
				start++;
			}
		}
		System.out.println();
		System.out.println(count);
		
	}
	
	private static Map<Long, Long> primeFactorization(long n, long m){
		Map<Long, Long> map = new HashMap<>();
		
		for(long i = 2; i <= Math.sqrt(n); i++){
			if(n % i == 0){
			long count = 0;
			while(n % i == 0){
				count++;
				n = n / i;
			}
			long val = count * m;
			map.put(i, val);
//			System.out.println("i: " + i + ", count: " + count);
		}
			
		}
		if(n != 1) {
//			System.out.println(n);
			map.put(n, m);
		}
		return map;
	}

	private static class Pair<T>{
		T a;
		T b;
		Pair(T a, T b){
			this.a = a;
			this.b = b;
		}
	}
	
	private static class SegmentTree{
		int n;
		private final int MAXN = 100007;
		private long[] tree = new long[MAXN << 2];
		private long[] lazy = new long[MAXN << 2];
		private long[] arr = new long[MAXN];
		
		
		/***
		 * 
		 * arr is 1 based index.
		 * 
		 * @param arr
		 * 
		 */
		SegmentTree(int n, long[] arr){
			this.n = n;
			for(int i = 0; i < arr.length; i++) {
				this.arr[i] = arr[i];
			}
		}
		
		void build(int index, int left, int right) {
			if(left == right) {
				tree[index] = arr[left];
			}
			else {
				int mid = (left + right) / 2;
				
				build(index * 2, left, mid);
				build((index * 2) + 1, mid + 1, right);
				tree[index] = max(tree[(index * 2)], tree[(index * 2) + 1]);
			}
		}
		
		long query(int node, int left, int right, int start, int end) {
			
			if(left > end || right < start) {
				return NEG_INF;
			}
			
			if(left >= start && right <= end) {
				return tree[node];
			}
			int mid = (left + right) / 2;
			long val1 = query(2 * node, left, mid, start, end);
			long val2 = query(2 * node + 1, mid + 1, right, start, end);
			
			return max(val1, val2);
		}
		
		void update(int node, int left, int right, int idx, long val) {
			if(left == right) {
				tree[node] += val;
			}
			else {
				int mid = (left + right) / 2;
				if(idx <= mid) {
					update(2 * node, left, mid, idx, val);
				}
				else {
					update(2 * node + 1, mid + 1, right, idx, val);
				}
				tree[node] = max(tree[(2 * node) + 1], tree[(2 * node)]);
			}
		}
		
		

		 void updateRange(int node, int start, int end, int l, int r, long val)
		{
		    if(lazy[node] != 0)
		    { 
		        // This node needs to be updated
		        tree[node] =  lazy[node];    // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];                  // Mark child as lazy
		            lazy[node*2+1] = lazy[node];                // Mark child as lazy
		        }
		        lazy[node] = 0;                                  // Reset it
		    }
		    if(start > end || start > r || end < l)              // Current segment is not within range [l, r]
		        return;
		    if(start >= l && end <= r)
		    {
		        // Segment is fully within range
		        tree[node] =  val;
		        if(start != end)
		        {
		            // Not leaf node
		            lazy[node*2] = val;
		            lazy[node*2+1] = val;
		        }
		        return;
		    }
		    int mid = (start + end) / 2;
		    updateRange(node*2, start, mid, l, r, val);        // Updating left child
		    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
		    tree[node] = max(tree[node*2], tree[node*2+1]);        // Updating root with max value 
		}

		 long queryRange(int node, int start, int end, int l, int r)
		{
		    if(start > end || start > r || end < l)
		        return 0;         // Out of range
		    if(lazy[node] != 0)
		    {
		        // This node needs to be updated
		        tree[node] =  lazy[node];            // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];         // Mark child as lazy
		            lazy[node*2+1] = lazy[node];    // Mark child as lazy
		        }
		        lazy[node] = 0;                 // Reset it
		    }
		    if(start >= l && end <= r)             // Current segment is totally within range [l, r]
		        return tree[node];
		    int mid = (start + end) / 2;
		    long p1 = queryRange(node*2, start, mid, l, r);         // Query left child
		    long p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
		    return max(p1, p2);
		}
		
		 void buildRange(int node, int low, int high){
			if(low == high){
				tree[node] =  arr[low];
				return;
			}
			
			int mid = (low + high) / 2;
			int left = node << 1;
			int right = left | 1;
			buildRange(left, low, mid);
			buildRange(right, mid + 1, high);
			tree[node] = max(tree[left], tree[right]);
		}
		
		
		void printSegmentTree() {
			System.out.println(Arrays.toString(tree));
		}
		
	}
	
	private static class KMP{
		
		private static char[] t;
		private static char[] s;
		
		public int kmp(char[]t, char[]s) {
			this.t = t;
			this.s = s;
			return this.kmp();
		}
		
		private int kmp() {
			
			List<Integer> prefixTable = getPrefixTable(s);
			
			int match = 0;
			int i = 0;
			int j = 0;
			
			int n = t.length;
			int m = s.length;
			while(i < n) {
				if(t[i] == s[j]) {
					if(j == m - 1) {
						match++;
						j = prefixTable.get(j - 1);
						continue;
					}
					i++;
					j++;
				}
				else if(j > 0) {
					j = prefixTable.get(j - 1);
				}
				else {
					i++;
				}
			}
			
			return match;
		}

		/***
		1. We compute the prefix values π[i] in a loop by iterating 	<br/>
				from i=1 to i=n−1 (π[0] just gets assigned with 0).		<br/><br/>
		2. To calculate the current value π[i] we set the variable j 	<br/>
				denoting the length of the best suffix for i−1. Initially j=π[i−1].	<br/>
		3. Test if the suffix of length j+1 is also a prefix by		 <br/><br/>
				comparing s[j] and s[i]. If they are equal then we assign π[i]=j+1, 	<br/>
				otherwise we reduce j to π[j−1] and repeat this step.		<br/><br/>
		4. If we have reached the length j=0 and still don't have a match,	<br/> 
				then we assign π[i]=0 and go to the next index i+1.		<br/><br/>
				
		@param
	 	pattern(String)
	***/
		private List<Integer> getPrefixTable(char[] pattern){
			
			List<Integer> prefixTable = new ArrayList<Integer>();
			int n = pattern.length;
			for(int i = 0; i < n; i++) {
				prefixTable.add(0);
			}
			
			for(int i = 1; i < n; i++) {
				for(int j = prefixTable.get(i - 1); j >= 0;) {
					if(pattern[j] == pattern[i]) {
						prefixTable.set(i, j + 1);
						break;
					}
					else if(j > 0){
						j = prefixTable.get(j - 1);
					}
					else {
						break;
					}
				}
			}
			
			return prefixTable;
		}
	}
	


	private static class Point {
		long x;
		long y;
		Point(long x, long y){
			this.x = x;
			this.y = y;
		}
		
		@Override
		public boolean equals(Object obj) {
			Point ob = (Point) obj;
			if(this.x == ob.x && this.y == ob.y){
				return true;
			}
			return false;
			
		}
		@Override
		public String toString() {
			return this.x + " " + this.y;
		}
		
		@Override
		public int hashCode() {
		    return 0;
		}
	}

	private static long pow(int base, int pow) {
		long val = 1L;
		for(int i = 1; i <= pow; i++) {
			val *= base;
		}
		return val;
	}
	
	private static int log(int x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static int log(long x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static long max(long a, long b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static long abs(long a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int abs(int a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int max(int a, int b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static int min(int a, int b) {
		if (a <= b) {
			return a;
		}
		return b;
	}
	
	private static long min(long a, long b) {
		if (a <= b) {
			return a;
		}
		return b;
	}

	private static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static long bigMod(long n, long k, long m) {

		long ans = 1;
		while (k > 0) {
			if ((k & 1) == 1) {
				ans = (ans * n) % m;
			}
			n = (n * n) % m;
			k >>= 1;
		}
		return ans;
	}

	/*
	 * Returns an iterator pointing to the first element in the range [first,
	 * last] which does not compare less than val.
	 * 
	 */
	private static int lowerBoundNew(long[] arr, long num) {
		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid >= 0 && arr[mid] == num) {
					mid--;
				}
				return mid + 1;
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	/*
	 * upper_bound() is a standard library function in C++ defined in the header
	 * . It returns an iterator pointing to the first element in the range
	 * [first, last) that is greater than value, or last if no such element is
	 * found
	 * 
	 */
	private static int upperBoundNew(long arr[], long num) {

		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			long val = arr[mid];
			if (val > num) {
				end = mid - 1;
			} else if (val < num) {
				start = mid + 1;
			} else {
				while (mid < len && arr[mid] == num) {
					mid++;
				}
				if (mid == len - 1 && arr[mid] == num) {
					return mid + 1;
				} else {
					return mid;
				}
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	private static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());

				}
			} catch (IOException e) {
				return null;
			}
			return tokenizer.nextToken();
		}

		public String nextLine() {
			String line = null;
			try {
				tokenizer = null;
				line = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return line;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean hasNext() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());
				}
			} catch (Exception e) {
				return false;
			}
			return true;
		}
	}
}