#include<bits/stdc++.h>
using namespace std;

const int _ = 4e5 + 7;
struct Edge{int end , upEd;}Ed[_ << 1];
int head[_] , cntEd , N , M , du[_]; char str[_];

void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd; ++du[b];}

int main(){
	scanf("%d %d %s" , &N , &M , str + 1);
	for(int i = 1 ; i <= M ; ++i){
		int x , y; scanf("%d %d" , &x , &y);
		if(str[x] == str[y]){addEd(x , y + N); addEd(y , x + N);}
		else{addEd(x + N , y); addEd(y + N , x);}
	}
	queue < int > q; int cnt = 0;
	for(int i = 1 ; i <= 2 * N ; ++i) if(!du[i]) q.push(i);
	while(!q.empty()){
		int t = q.front(); q.pop(); ++cnt;
		for(int i = head[t] ; i ; i = Ed[i].upEd)
			if(!--du[Ed[i].end]) q.push(Ed[i].end);
	}
	puts(cnt == 2 * N ? "No" : "Yes"); return 0;
}
### 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>
using namespace std;

const int _ = 4e5 + 7;
struct Edge{int end , upEd;}Ed[_ << 1];
int head[_] , cntEd , N , M , du[_]; char str[_];

void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd; ++du[b];}

int main(){
	scanf("%d %d %s" , &N , &M , str + 1);
	for(int i = 1 ; i <= M ; ++i){
		int x , y; scanf("%d %d" , &x , &y);
		if(str[x] == str[y]){addEd(x , y + N); addEd(y , x + N);}
		else{addEd(x + N , y); addEd(y + N , x);}
	}
	queue < int > q; int cnt = 0;
	for(int i = 1 ; i <= 2 * N ; ++i) if(!du[i]) q.push(i);
	while(!q.empty()){
		int t = q.front(); q.pop(); ++cnt;
		for(int i = head[t] ; i ; i = Ed[i].upEd)
			if(!--du[Ed[i].end]) q.push(Ed[i].end);
	}
	puts(cnt == 2 * N ? "No" : "Yes"); return 0;
}
```