#include <cstdio>
#include <queue>

const int N=2e5+5,M=4e5+5;
int n,m,tot,lnk[N],ter[M],nxt[M],deg[N][2];
char col[N];
bool vis[N];

void add(int u,int v) {
	ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot;
}
bool check() {
	for(int i=1;i<=n;++i) if(!vis[i]) return 0;
	return 1;
}
int main() {
	scanf("%d%d%s",&n,&m,col+1);
	while(m--) {
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v),add(v,u);
		++deg[u][col[v]=='B'];
		++deg[v][col[u]=='B'];
	}
	std::queue<int> q;
	for(int i=1;i<=n;++i) {
		if(!deg[i][0]||!deg[i][1]) q.push(i),vis[i]=1;
	}
	while(!q.empty()) {
		int u=q.front(); q.pop();
		for(int i=lnk[u];i;i=nxt[i]) {
			int v=ter[i];
			if(vis[v]) continue;
			if(!--deg[v][col[u]=='B']) q.push(v),vis[v]=1;
		}

	}
	puts(check()?"No":"Yes");
	return 0;
}
### Prompt

Generate a cpp solution to the following problem:

You are given an undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M. In addition, each vertex has a label, `A` or `B`. The label of Vertex i is s_i. Edge i bidirectionally connects vertex a_i and b_i.

The phantom thief Nusook likes to choose some vertex as the startpoint and traverse an edge zero or more times. Today, he will make a string after traveling as above, by placing the labels of the visited vertices in the order visited, beginning from the startpoint.

For example, in a graph where Vertex 1 has the label `A` and Vertex 2 has the label `B`, if Nusook travels along the path 1 \rightarrow 2 \rightarrow 1 \rightarrow 2 \rightarrow 2, the resulting string is `ABABB`.

Determine if Nusook can make all strings consisting of `A` and `B`.

Constraints

* 2 \leq N \leq 2 \times 10^{5}
* 1 \leq M \leq 2 \times 10^{5}
* |s| = N
* s_i is `A` or `B`.
* 1 \leq a_i, b_i \leq N
* The given graph may NOT be simple or connected.

Input

Input is given from Standard Input in the following format:


N M
s
a_1 b_1
:
a_{M} b_{M}


Output

If Nusook can make all strings consisting of `A` and `B`, print `Yes`; otherwise, print `No`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

13 23
ABAAAABBBBAAB
7 1
10 6
1 11
2 10
2 8
2 11
11 12
8 3
7 12
11 2
13 13
11 9
4 1
9 7
9 6
8 13
8 6
4 10
8 7
4 3
2 1
8 12
6 9


Output

Yes


Input

13 17
BBABBBAABABBA
7 1
7 9
11 12
3 9
11 9
2 1
11 5
12 11
10 8
1 11
1 8
7 7
9 10
8 8
8 12
6 2
13 11


Output

No

### Response

```cpp
#include <cstdio>
#include <queue>

const int N=2e5+5,M=4e5+5;
int n,m,tot,lnk[N],ter[M],nxt[M],deg[N][2];
char col[N];
bool vis[N];

void add(int u,int v) {
	ter[++tot]=v,nxt[tot]=lnk[u],lnk[u]=tot;
}
bool check() {
	for(int i=1;i<=n;++i) if(!vis[i]) return 0;
	return 1;
}
int main() {
	scanf("%d%d%s",&n,&m,col+1);
	while(m--) {
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v),add(v,u);
		++deg[u][col[v]=='B'];
		++deg[v][col[u]=='B'];
	}
	std::queue<int> q;
	for(int i=1;i<=n;++i) {
		if(!deg[i][0]||!deg[i][1]) q.push(i),vis[i]=1;
	}
	while(!q.empty()) {
		int u=q.front(); q.pop();
		for(int i=lnk[u];i;i=nxt[i]) {
			int v=ter[i];
			if(vis[v]) continue;
			if(!--deg[v][col[u]=='B']) q.push(v),vis[v]=1;
		}

	}
	puts(check()?"No":"Yes");
	return 0;
}
```