#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
using namespace std;
int n,m,De[2020202][2];
char S[2020202];
bool Out[2020202];
vector <int> Vec[202020];
stack <int> St;
int main(){
	scanf("%d %d",&n,&m);
	scanf("%s",S+1);
	for (int i=1;i<=m;i++){
		int x,y; scanf("%d %d",&x,&y);
		De[x][S[y]-'A']++; De[y][S[x]-'A']++;
		Vec[x].push_back(y);
		Vec[y].push_back(x);
	}
	for (int i=1;i<=n;i++) if (!De[i][0]||!De[i][1])
		St.push(i),Out[i]=1;;
	while (!St.empty()){
		int x=St.top(); St.pop(); 
		for (int i=0;i<Vec[x].size();i++){
			int y=Vec[x][i];
			if (Out[y]) continue;
			De[y][S[x]-'A']--;
			if (!De[y][0]||!De[y][1]){
				Out[y]=1; St.push(y);
			}
		}
	}
	for (int i=1;i<=n;i++) if (!Out[i]){
		cout<<"Yes"; return 0;
	}
	cout<<"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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
using namespace std;
int n,m,De[2020202][2];
char S[2020202];
bool Out[2020202];
vector <int> Vec[202020];
stack <int> St;
int main(){
	scanf("%d %d",&n,&m);
	scanf("%s",S+1);
	for (int i=1;i<=m;i++){
		int x,y; scanf("%d %d",&x,&y);
		De[x][S[y]-'A']++; De[y][S[x]-'A']++;
		Vec[x].push_back(y);
		Vec[y].push_back(x);
	}
	for (int i=1;i<=n;i++) if (!De[i][0]||!De[i][1])
		St.push(i),Out[i]=1;;
	while (!St.empty()){
		int x=St.top(); St.pop(); 
		for (int i=0;i<Vec[x].size();i++){
			int y=Vec[x][i];
			if (Out[y]) continue;
			De[y][S[x]-'A']--;
			if (!De[y][0]||!De[y][1]){
				Out[y]=1; St.push(y);
			}
		}
	}
	for (int i=1;i<=n;i++) if (!Out[i]){
		cout<<"Yes"; return 0;
	}
	cout<<"No";
}
```