Andrewid the Android is a galaxy-known detective. Now he is preparing a defense against a possible attack by hackers on a major computer network.

In this network are n vertices, some pairs of vertices are connected by m undirected channels. It is planned to transfer q important messages via this network, the i-th of which must be sent from vertex si to vertex di via one or more channels, perhaps through some intermediate vertices.

To protect against attacks a special algorithm was developed. Unfortunately it can be applied only to the network containing directed channels. Therefore, as new channels can't be created, it was decided for each of the existing undirected channels to enable them to transmit data only in one of the two directions.

Your task is to determine whether it is possible so to choose the direction for each channel so that each of the q messages could be successfully transmitted.

Input

The first line contains three integers n, m and q (1 ≤ n, m, q ≤ 2·105) — the number of nodes, channels and important messages.

Next m lines contain two integers each, vi and ui (1 ≤ vi, ui ≤ n, vi ≠ ui), that means that between nodes vi and ui is a channel. Between a pair of nodes can exist more than one channel.

Next q lines contain two integers si and di (1 ≤ si, di ≤ n, si ≠ di) — the numbers of the nodes of the source and destination of the corresponding message.

It is not guaranteed that in it initially possible to transmit all the messages.

Output

If a solution exists, print on a single line "Yes" (without the quotes). Otherwise, print "No" (without the quotes).

Examples

Input

4 4 2
1 2
1 3
2 3
3 4
1 3
4 2


Output

Yes


Input

3 2 2
1 2
3 2
1 3
2 1


Output

No


Input

3 3 2
1 2
1 2
3 2
1 3
2 1


Output

Yes

Note

In the first sample test you can assign directions, for example, as follows: 1 → 2, 1 → 3, 3 → 2, 4 → 3. Then the path for for the first message will be 1 → 3, and for the second one — 4 → 3 → 2.

In the third sample test you can assign directions, for example, as follows: 1 → 2, 2 → 1, 2 → 3. Then the path for the first message will be 1 → 2 → 3, and for the second one — 2 → 1.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;
public class CaseOfComputerNetwork {
	static ArrayList<Edge>[] graph;
	static int[] ll;
	static int[] in;
	static int[] col;
	static int[] comp;
	static boolean[] vis;
	static int cnt = 0;
	static Edge[] edges;
	static int currcolor = -1;
	static ArrayList<Integer>[] aux;
	static int[][] bl;
	static int blsz;
	static int[] depths;
	static int[] preflca;
	static int[] prefstart;
	static int[] prefend;
	static int[] nstarts;
	static int[] nends;
	static int[] lcas;
	public static void main(String[] args) {
		FastScanner scanner = new FastScanner();
		int N = scanner.nextInt();
		int M = scanner.nextInt();
		int Q = scanner.nextInt();
		graph = new ArrayList[N];
		for(int i = 0; i < N;i++) graph[i] = new ArrayList<>();
		edges = new Edge[M];
		for(int i = 0; i < M; i++) {
			int u = scanner.nextInt()-1;
			int v = scanner.nextInt()-1;
			edges[i] = new Edge(u, v);
			graph[u].add(edges[i]);
			graph[v].add(edges[i]);
		}
		vis = new boolean[N];
		in = new int[N];
		col = new int[N];
		comp = new int[N];
		ll = new int[N];
		Arrays.fill(comp, -1);
		Arrays.fill(in,  -1);
		Arrays.fill(ll,  -1);
		Arrays.fill(col,  -1);
		int curCol = 0;
		for(int i =0; i < N; i++) {
			if (in[i] == -1) {
				findBridges(i, null, curCol++);
			}
		}
		for(int i = 0; i < N; i++) {
			if (col[i] == -1) {
				color(i,null, ++currcolor);
			}
		}
		currcolor++;
		aux = new ArrayList[currcolor];
		for(int i = 0; i < currcolor; i++) aux[i] = new ArrayList<>();
		for(int i = 0; i < M; i++) {
			if (edges[i].isBridge) {
				aux[col[edges[i].u]].add(col[edges[i].v]);
				aux[col[edges[i].v]].add(col[edges[i].u]);
			}
		}
		depths = new int[currcolor];
		Arrays.fill(depths,  -1);
		ArrayDeque<Integer> queue = new ArrayDeque<>();
		blsz = Integer.numberOfTrailingZeros(Integer.highestOneBit(currcolor));
		bl = new int[currcolor][blsz + 1];
		for(int i = 0; i < currcolor; i++) {
			if (depths[i] == -1) {
				queue.offer(i);
				depths[i] = 0;
				while(!queue.isEmpty()) {
					int cur = queue.poll();
					for(int edge: aux[cur]) {
						if (depths[edge] >= 0) continue;
						depths[edge] = depths[cur]+1;
						bl[edge][0] = cur;
						queue.offer(edge);
					}
				}
			}
		}
		for(int i = 1; i <= blsz; i++) {
			for(int j = 0; j < currcolor; j++) {
				bl[j][i] = bl[bl[j][i-1]][i-1];
			}
		}
		int[] starts = new int[Q];
		int[] ends = new int[Q];
		nstarts = new int[currcolor];
		nends = new int[currcolor];
		lcas = new int[currcolor];
		preflca = new int[currcolor];
		prefstart = new int[currcolor];
		prefend = new int[currcolor];
		for(int i = 0; i < Q; i++) {
			starts[i] = scanner.nextInt()-1;
			ends[i] = scanner.nextInt()-1;
			if (comp[starts[i]] != comp[ends[i]]) {
				System.out.println("No");
				return;
			}
			int curlca = lca(col[starts[i]], col[ends[i]]);
			nstarts[col[starts[i]]]++;
			nends[col[ends[i]]]++;
			lcas[curlca]++;
		}
		for(int i = 0; i < currcolor; i++) {
			if (depths[i] == 0) getAns(i, -1);
		}
		System.out.println("Yes");
	}
	
	static void getAns(int cur, int p) {
		//todo need to make sure that the number of starts/ends in some path
		//does not create an issue.
		//ie need to make sure that we correctly close paths when all the corresponding ends have
		//been found and need to make sure that we correctly verify that that we have overlapping
		//paths that go in opposite directions.
		preflca[cur] += lcas[cur];
		prefend[cur] += nends[cur];
		prefstart[cur] += nstarts[cur];
		for(int edge: aux[cur]) {
			if (edge == p) continue;
			getAns(edge, cur);
			preflca[cur] += preflca[edge];
			prefend[cur] += prefend[edge];
			prefstart[cur] += prefstart[edge];
		}
		if (prefstart[cur] - preflca[cur] > 0 && prefend[cur] - preflca[cur] > 0) {
			System.out.println("No");
			System.exit(0);
		}
	}
	
	static int lca(int u, int v) {
		if (depths[u] > depths[v]) {
			u = walk(u, depths[u] - depths[v]);
		}
		if (depths[v] > depths[u]) {
			v = walk(v, depths[v] - depths[u]);
		}
		if (u == v) return u;
		for(int i = blsz; i >= 0; i--) {
			if (bl[u][i] != bl[v][i]) {
				u = bl[u][i];
				v = bl[v][i];
			}
		}
		return bl[u][0];
	}
	
	static int walk(int v, int d) {
		for(int i = 0; i <= blsz; i++) {
			if (((1 << i) & d) > 0) v = bl[v][i];
		}
		return v;
	}
	
	static void color(int v, Edge prev, int curCol) {
		vis[v] = true;
		col[v] = curCol;
		for(Edge edge: graph[v]) {
			int u = edge.getOther(v);
			if (edge == prev || vis[u]) continue;
			if (edge.isBridge) color(u, edge, ++currcolor);
			else color(u, edge, curCol);
		}
	}
	
	static void findBridges(int v, Edge prev, int curCol) {
		comp[v] = curCol;
		in[v] = ll[v] = cnt++;
		for(Edge edge: graph[v]) {
			int u = edge.getOther(v);
			if (edge == prev) continue;
			if (in[u] != -1) {
				ll[v] = Math.min(ll[v],  in[u]);
			}
			else {
				findBridges(u, edge, curCol);
				ll[v] = Math.min(ll[v],  ll[u]);
			}
			if (ll[u] > in[v]) {
				edge.isBridge = true;
			}
		}
	}
	
	static class Edge {
		int u, v;
		boolean isBridge;
		public Edge(int uu, int vv) {
			u = uu; v = vv;
		}
		int getOther(int x) {
			if (x == u) return v;
			return u;
		}
	}
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner(Reader in) {
			br = new BufferedReader(in);
		}
		
		public FastScanner() {
			this(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
		
		String readNextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
		
		int[] readIntArray(int n) {
			int[] a = new int[n];
			for (int idx = 0; idx < n; idx++) {
				a[idx] = nextInt();
			}
			return a;
		}
	}
}
