#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int N,M;
int A[SIZE],B[SIZE];
vector<int> G[SIZE];
string S;

int main()
{
	cin >> N >> M >> S;
	rep(i,0,M){
		int a,b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
		if(S[b-1]=='A'){
			A[a]++;
		}else{
			B[a]++;
		}
		if(S[a-1]=='A'){
			A[b]++;
		}else{
			B[b]++;
		}
	}
	int cnt = 0;

	queue<int> Q;
	rep(i,1,N+1){
		if(A[i]==0 || B[i]==0){
			Q.push(i);
			cnt++;
		}
	}

	while(!Q.empty()){
		int v = Q.front();Q.pop();
		for(int c : G[v]){
			if(S[v-1]=='A'){
				A[c]--;
				if(A[c]==0 && B[c]>0){
					Q.push(c);
					cnt++;
				}
			}else{
				B[c]--;
				if(B[c]==0 && A[c]>0){
					Q.push(c);
					cnt++;
				}
			}
		}
	}
	if(N != cnt){
		cout << "Yes" << endl;
	}else{
		cout << "No" << endl;
	}

	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 200005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int N,M;
int A[SIZE],B[SIZE];
vector<int> G[SIZE];
string S;

int main()
{
	cin >> N >> M >> S;
	rep(i,0,M){
		int a,b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
		if(S[b-1]=='A'){
			A[a]++;
		}else{
			B[a]++;
		}
		if(S[a-1]=='A'){
			A[b]++;
		}else{
			B[b]++;
		}
	}
	int cnt = 0;

	queue<int> Q;
	rep(i,1,N+1){
		if(A[i]==0 || B[i]==0){
			Q.push(i);
			cnt++;
		}
	}

	while(!Q.empty()){
		int v = Q.front();Q.pop();
		for(int c : G[v]){
			if(S[v-1]=='A'){
				A[c]--;
				if(A[c]==0 && B[c]>0){
					Q.push(c);
					cnt++;
				}
			}else{
				B[c]--;
				if(B[c]==0 && A[c]>0){
					Q.push(c);
					cnt++;
				}
			}
		}
	}
	if(N != cnt){
		cout << "Yes" << endl;
	}else{
		cout << "No" << endl;
	}

	return 0;
}

```