#include <bits/stdc++.h>
using namespace std;
#define N 200100
int n,m,he[N],ne[N<<1],e[N<<1],cnt,fa[N],op[N],C[N][2],Q[N];
char s[N];
void add(int u,int v){ne[++cnt]=he[u];he[u]=cnt;e[cnt]=v;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=1;i<=m;i++){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
		C[u][s[v]-'A']++; C[v][s[u]-'A']++;
	}
	int h=0,t=0;
	for (int i=1;i<=n;i++) if (!C[i][0] || !C[i][1]) Q[++t]=i;
	while (h!=t){
		int x=Q[++h];
		for (int i=he[x];i;i=ne[i]) {
			C[e[i]][s[x]-'A']--;
			if (!C[e[i]][s[x]-'A'] && C[e[i]][(s[x]-'A')^1]) Q[++t]=e[i];
		}
	}
	if (t==n) puts("No"); else puts("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 <bits/stdc++.h>
using namespace std;
#define N 200100
int n,m,he[N],ne[N<<1],e[N<<1],cnt,fa[N],op[N],C[N][2],Q[N];
char s[N];
void add(int u,int v){ne[++cnt]=he[u];he[u]=cnt;e[cnt]=v;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=1;i<=m;i++){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v); add(v,u);
		C[u][s[v]-'A']++; C[v][s[u]-'A']++;
	}
	int h=0,t=0;
	for (int i=1;i<=n;i++) if (!C[i][0] || !C[i][1]) Q[++t]=i;
	while (h!=t){
		int x=Q[++h];
		for (int i=he[x];i;i=ne[i]) {
			C[e[i]][s[x]-'A']--;
			if (!C[e[i]][s[x]-'A'] && C[e[i]][(s[x]-'A')^1]) Q[++t]=e[i];
		}
	}
	if (t==n) puts("No"); else puts("Yes");
	return 0;
}

```