//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std; typedef long long ll;
inline int qr(){
	int ret=0,f=0,c=getchar();
	while(!isdigit(c)) f|=c==45,c=getchar();
	while( isdigit(c)) ret=ret*10+c-48,c=getchar();
	return f?-ret:ret;
}
const int maxn=2e5+5;
int n,m;
char w[maxn];
vector<int> e[maxn<<1];
int usd[maxn<<1];
void add(int fr,int to){e[fr].push_back(to);}

void dfs(int now,int last){
	usd[now]=1;
	for(auto t:e[now])
		if(t^last){
			if(!usd[t]) dfs(t,now);
			else if(usd[t]!=-1) puts("Yes"),exit(0);
		}
	usd[now]=-1;
}

int main(){
	n=qr(); m=qr();
	scanf("%s",w+1);
	for(int t=1,a,b;t<=m;++t){
		a=qr(),b=qr();
		if(w[a]!=w[b]) add(a,b+n),add(b,a+n);
		else add(a+n,b),add(b+n,a);
	}
	for(int t=1;t<=n+n;++t)
		if(!usd[t]) dfs(t,0);
	puts("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
//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>

using namespace std; typedef long long ll;
inline int qr(){
	int ret=0,f=0,c=getchar();
	while(!isdigit(c)) f|=c==45,c=getchar();
	while( isdigit(c)) ret=ret*10+c-48,c=getchar();
	return f?-ret:ret;
}
const int maxn=2e5+5;
int n,m;
char w[maxn];
vector<int> e[maxn<<1];
int usd[maxn<<1];
void add(int fr,int to){e[fr].push_back(to);}

void dfs(int now,int last){
	usd[now]=1;
	for(auto t:e[now])
		if(t^last){
			if(!usd[t]) dfs(t,now);
			else if(usd[t]!=-1) puts("Yes"),exit(0);
		}
	usd[now]=-1;
}

int main(){
	n=qr(); m=qr();
	scanf("%s",w+1);
	for(int t=1,a,b;t<=m;++t){
		a=qr(),b=qr();
		if(w[a]!=w[b]) add(a,b+n),add(b,a+n);
		else add(a+n,b),add(b+n,a);
	}
	for(int t=1;t<=n+n;++t)
		if(!usd[t]) dfs(t,0);
	puts("No");
	return 0;
}

```