#include <bits/stdc++.h>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define for1(a,b,i) for(int i=a;i<=b;++i)
#define FOR2(a,b,i) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
inline int read() {
	int f=1,sum=0;
	char x=getchar();
	for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1;
	for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0';
	return f*sum;
}

#define M 1000005
int n,m;
char s[M];
int e_size,head[M],vis[M];

struct node {int v,nxt;}e[M*2];

inline void e_add(int u,int v) {
	e[++e_size]=(node){v,head[u]};
	head[u]=e_size;
}

inline void dfs(int x) {
	vis[x]=1;
	for(int i=head[x];i;i=e[i].nxt) {
		int v=e[i].v;
		if(vis[v]==1) {
			puts("Yes");
			exit(0);
		}
		if(vis[v]==0) dfs(v);
 	}
 	vis[x]=-1;
}

int main () {
//	freopen("a.in","r",stdin);
	n=read(),m=read();
	scanf("%s",s+1);
	for1(1,m,i) {
		int x=read(),y=read();
		if(s[x]==s[y]) e_add(x,y+n),e_add(y,x+n);
		else e_add(x+n,y),e_add(y+n,x);
	}
	for1(1,n,i) if(!vis[i]) dfs(i);
	puts("No");
}
### Prompt

In CPP, your task is to solve 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 min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define for1(a,b,i) for(int i=a;i<=b;++i)
#define FOR2(a,b,i) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
inline int read() {
	int f=1,sum=0;
	char x=getchar();
	for(;(x<'0'||x>'9');x=getchar()) if(x=='-') f=-1;
	for(;x>='0'&&x<='9';x=getchar()) sum=sum*10+x-'0';
	return f*sum;
}

#define M 1000005
int n,m;
char s[M];
int e_size,head[M],vis[M];

struct node {int v,nxt;}e[M*2];

inline void e_add(int u,int v) {
	e[++e_size]=(node){v,head[u]};
	head[u]=e_size;
}

inline void dfs(int x) {
	vis[x]=1;
	for(int i=head[x];i;i=e[i].nxt) {
		int v=e[i].v;
		if(vis[v]==1) {
			puts("Yes");
			exit(0);
		}
		if(vis[v]==0) dfs(v);
 	}
 	vis[x]=-1;
}

int main () {
//	freopen("a.in","r",stdin);
	n=read(),m=read();
	scanf("%s",s+1);
	for1(1,m,i) {
		int x=read(),y=read();
		if(s[x]==s[y]) e_add(x,y+n),e_add(y,x+n);
		else e_add(x+n,y),e_add(y+n,x);
	}
	for1(1,n,i) if(!vis[i]) dfs(i);
	puts("No");
}
```