#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
const int N=2e5+7;
int num[N][2],n,m,die[N];
vector<int> tu[N];
char s[N];
queue<int> Q;
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	rep(i,1,m){
		int u,v;scanf("%d%d",&u,&v);
		tu[u].push_back(v);
		tu[v].push_back(u);
		num[u][s[v]-'A']++;
		num[v][s[u]-'A']++;
	}
	rep(i,1,n)if(!num[i][0] || !num[i][1])Q.push(i);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		if(die[u])continue;
		die[u]=1;
		for(int i=0;i<tu[u].size();i++){
			int v=tu[u][i];
			num[v][s[u]-'A']--;
			if(!num[v][s[u]-'A'])Q.push(v);
		}
	}
	int ok=0;
	rep(i,1,n)if(!die[i])ok=1;
	if(ok)puts("Yes");
	else puts("No");
}
### Prompt

Your task is to create 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>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define rep2(i,k,n) for(int i=k;i>=n;i--)
using namespace std;
const int N=2e5+7;
int num[N][2],n,m,die[N];
vector<int> tu[N];
char s[N];
queue<int> Q;
int main(){
	scanf("%d%d%s",&n,&m,s+1);
	rep(i,1,m){
		int u,v;scanf("%d%d",&u,&v);
		tu[u].push_back(v);
		tu[v].push_back(u);
		num[u][s[v]-'A']++;
		num[v][s[u]-'A']++;
	}
	rep(i,1,n)if(!num[i][0] || !num[i][1])Q.push(i);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		if(die[u])continue;
		die[u]=1;
		for(int i=0;i<tu[u].size();i++){
			int v=tu[u][i];
			num[v][s[u]-'A']--;
			if(!num[v][s[u]-'A'])Q.push(v);
		}
	}
	int ok=0;
	rep(i,1,n)if(!die[i])ok=1;
	if(ok)puts("Yes");
	else puts("No");
}
```