#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

using graph=vector<vector<int>>;

int main(){
	int n,m;
	string s; cin>>n>>m>>s;
	graph G(n);
	rep(i,m){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		G[u].emplace_back(v);
		if(u!=v) G[v].emplace_back(u);
	}

	vector<vector<int>> deg(2,vector<int>(n));
	rep(u,n) for(int v:G[u]) deg[s[v]-'A'][u]++;

	vector<bool> del(n);
	queue<int> Q;
	rep(u,n) if(deg[0][u]==0 || deg[1][u]==0) {
		del[u]=true;
		Q.emplace(u);
	}
	while(!Q.empty()){
		int u=Q.front(); Q.pop();
		for(int v:G[u]) if(!del[v]) {
			deg[s[u]-'A'][v]--;
			if(deg[0][v]==0 || deg[1][v]==0){
				del[v]=true;
				Q.emplace(v);
			}
		}
	}

	puts(count(del.begin(),del.end(),false)>0?"Yes":"No");

	return 0;
}

### Prompt

Create a solution in Cpp for 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,n) for(int i=0;i<(n);i++)

using namespace std;

using graph=vector<vector<int>>;

int main(){
	int n,m;
	string s; cin>>n>>m>>s;
	graph G(n);
	rep(i,m){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		G[u].emplace_back(v);
		if(u!=v) G[v].emplace_back(u);
	}

	vector<vector<int>> deg(2,vector<int>(n));
	rep(u,n) for(int v:G[u]) deg[s[v]-'A'][u]++;

	vector<bool> del(n);
	queue<int> Q;
	rep(u,n) if(deg[0][u]==0 || deg[1][u]==0) {
		del[u]=true;
		Q.emplace(u);
	}
	while(!Q.empty()){
		int u=Q.front(); Q.pop();
		for(int v:G[u]) if(!del[v]) {
			deg[s[u]-'A'][v]--;
			if(deg[0][v]==0 || deg[1][v]==0){
				del[v]=true;
				Q.emplace(v);
			}
		}
	}

	puts(count(del.begin(),del.end(),false)>0?"Yes":"No");

	return 0;
}

```