Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:

  * for the array [0, 0, 1, 0, 2] MEX equals to 3 because numbers 0, 1 and 2 are presented in the array and 3 is the minimum non-negative integer not presented in the array; 
  * for the array [1, 2, 3, 4] MEX equals to 0 because 0 is the minimum non-negative integer not presented in the array; 
  * for the array [0, 1, 4, 3] MEX equals to 2 because 2 is the minimum non-negative integer not presented in the array. 



You are given an empty array a=[] (in other words, a zero-length array). You are also given a positive integer x.

You are also given q queries. The j-th query consists of one integer y_j and means that you have to append one element y_j to the array. The array length increases by 1 after a query.

In one move, you can choose any index i and set a_i := a_i + x or a_i := a_i - x (i.e. increase or decrease any element of the array by x). The only restriction is that a_i cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of q queries (i.e. the j-th answer corresponds to the array of length j).

Operations are discarded before each query. I.e. the array a after the j-th query equals to [y_1, y_2, ..., y_j].

Input

The first line of the input contains two integers q, x (1 ≤ q, x ≤ 4 ⋅ 10^5) — the number of queries and the value of x.

The next q lines describe queries. The j-th query consists of one integer y_j (0 ≤ y_j ≤ 10^9) and means that you have to append one element y_j to the array.

Output

Print the answer to the initial problem after each query — for the query j print the maximum value of MEX after first j queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.

Examples

Input


7 3
0
1
2
2
0
0
10


Output


1
2
3
3
4
4
7


Input


4 3
1
2
1
2


Output


0
0
0
0

Note

In the first example:

  * After the first query, the array is a=[0]: you don't need to perform any operations, maximum possible MEX is 1. 
  * After the second query, the array is a=[0, 1]: you don't need to perform any operations, maximum possible MEX is 2. 
  * After the third query, the array is a=[0, 1, 2]: you don't need to perform any operations, maximum possible MEX is 3. 
  * After the fourth query, the array is a=[0, 1, 2, 2]: you don't need to perform any operations, maximum possible MEX is 3 (you can't make it greater with operations). 
  * After the fifth query, the array is a=[0, 1, 2, 2, 0]: you can perform a[4] := a[4] + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3]. Now MEX is maximum possible and equals to 4. 
  * After the sixth query, the array is a=[0, 1, 2, 2, 0, 0]: you can perform a[4] := a[4] + 3 = 0 + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3, 0]. Now MEX is maximum possible and equals to 4. 
  * After the seventh query, the array is a=[0, 1, 2, 2, 0, 0, 10]. You can perform the following operations: 
    * a[3] := a[3] + 3 = 2 + 3 = 5, 
    * a[4] := a[4] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 3 + 3 = 6, 
    * a[6] := a[6] - 3 = 10 - 3 = 7, 
    * a[6] := a[6] - 3 = 7 - 3 = 4. 
The resulting array will be a=[0, 1, 2, 5, 3, 6, 4]. Now MEX is maximum possible and equals to 7. 
import java.io.*;
import java.math.*;
import java.util.*;

/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { -2, -3, -2, -1, -1, 1};
	private static int dy[] = { 1, -1, -1, -2, -3, -2};

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

	private static final int MAX = 1007;
	private static final int MOD = 10007;
	
	private static final int MAXN = 400003;
	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);
		

/*



*/
//		System.out.println((4 % 1));
		int test = in.nextInt();
		int x = in.nextInt();
		
		SegmentTree segmentTree = new SegmentTree(x);
		segmentTree.build(1, 1, x);
//		segmentTree.printSegmentTree();
		for(int t = 1; t <= test; t++) {
			int a = in.nextInt();
			int rem = a % x;
			if(rem == 0) {
				rem = x;
			}
			segmentTree.update(1, 1, x, rem, 1);
//			segmentTree.printSegmentTree();
			Point point = segmentTree.tree[1];
//			System.out.println(point);
			rem = point.y;
			if(point.y == x) {
				rem = 0;
			}
			int ans = (point.x * x) + rem;
			out.println(ans);
		}
		
		
		out.flush();
		out.close();
		System.exit(0);
	}

	
	static Map<Integer, Integer> primeFactorization(long n){
		Map<Integer, Integer> map = new HashMap<Integer, Integer>();
		
		for(int i = 2; i <= Math.sqrt(n); i++){
			if(n % i == 0){
			int count = 0;
			while(n % i == 0){
				count++;
				n = n / i;
			}
			map.put(i, count);
//			System.out.println("i: " + i + ", count: " + count);
		}
			
		}
		if(n != 1) {
//			System.out.println(n);
			map.put((int)n, 1);
		}
		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 int MAXN = 20;
		private Point[] tree;// = new Point[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];
			}
		}
		
		SegmentTree(int n){
			this.n = n;
			this.MAXN = n;
			tree = new Point[MAXN << 2];
//			for(int i = 0; i < tree.length; i++) {
//				tree[i] = new Point(0, -1);
//			}
		}
		
		void build(int index, int left, int right) {
			if(left == right) {
				tree[index] = new Point(0, left);
			}
			else {
				int mid = (left + right) / 2;
				
				build(index * 2, left, mid);
				build((index * 2) + 1, mid + 1, right);
//				tree[index] = min(tree[(index * 2)], tree[(index * 2) + 1]);
				if(tree[(index * 2)].x < tree[(index * 2) + 1].x) {
					tree[index] = tree[(index * 2)];
				}
				else if(tree[(index * 2)].x == tree[(index * 2) + 1].x) {
					if(tree[(index * 2) + 1].y == n) {
						tree[index] = tree[(index * 2) + 1]; 
					}
					else {
						tree[index] = tree[index * 2];
					}
				}
				else {
					tree[index] = tree[(index * 2) + 1];
				}
			}
		}
		
		Point query(int node, int left, int right, int start, int end) {
			
			if(left > end || right < start) {
				return new Point(MAXN, -1);
			}
			
			if(left >= start && right <= end) {
				return tree[node];
			}
			int mid = (left + right) / 2;
			Point val1 = query(2 * node, left, mid, start, end);
			Point val2 = query(2 * node + 1, mid + 1, right, start, end);
			if(val1.x < val2.x) {
				return val1;
			}
			else if(val1.x == val2.x) {
				if(tree[(2 * node) + 1].y == n) {
					return val2;
				}
				return val1;
			}
			else {
				return val2;
			}
		}
		
		void update(int node, int left, int right, int idx, long val) {
			if(left == right) {
				tree[node].x += 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] = min(tree[(2 * node) + 1], tree[(2 * node)]);
				if(tree[2 * node].x < tree[(2 * node) + 1].x) {
					tree[node] = tree[2 * node];
				}
				else if(tree[(2 * node)].x == tree[(2 * node) + 1].x) {
					if(tree[(2 * node) + 1].y == n) {
						tree[node] = tree[(2 * node) + 1];
					}
					else {
						tree[node] = tree[2 * node];
					}
				}
				else {
					tree[node] = tree[(2 * node) + 1];
				}
			}
		}
		
		
/*
		 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 {
		int x;
		int y;
		Point(int x, int 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;
		}
	}

	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;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < 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;
		}
	}
}