#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=200010;
int tt;
char w[N],s[N];
int head[N],to[N*2],nxt[N*2],v[N],in[N][2];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=='-'?o=-1:0,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return o*x;
}
int main() {
    int n,m,tp=0;
    cin>>n>>m;
    scanf("%s",s+1);
    for(int i=1;i<=m;i++) {
	int x=gi(),y=gi();
	++in[y][s[x]-'A'],++in[x][s[y]-'A'];
	to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
    }
    for(int i=1;i<=n;i++)
	if(!in[i][0]||!in[i][1]) v[++tp]=i,w[i]=1;
    while(tp) {
	int x=v[tp--];
	for(int i=head[x];i;i=nxt[i]) {
	    int y=to[i];
	    if(!w[y]&&!--in[y][s[x]-'A']) v[++tp]=y,w[y]=1;
	}
    }
    for(int i=1;i<=n;i++)
	if(!w[i]) return puts("Yes"),0;
    puts("No");
    return 0;
}
### Prompt

Please create a solution in CPP 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 <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=200010;
int tt;
char w[N],s[N];
int head[N],to[N*2],nxt[N*2],v[N],in[N][2];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') ch=='-'?o=-1:0,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return o*x;
}
int main() {
    int n,m,tp=0;
    cin>>n>>m;
    scanf("%s",s+1);
    for(int i=1;i<=m;i++) {
	int x=gi(),y=gi();
	++in[y][s[x]-'A'],++in[x][s[y]-'A'];
	to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
    }
    for(int i=1;i<=n;i++)
	if(!in[i][0]||!in[i][1]) v[++tp]=i,w[i]=1;
    while(tp) {
	int x=v[tp--];
	for(int i=head[x];i;i=nxt[i]) {
	    int y=to[i];
	    if(!w[y]&&!--in[y][s[x]-'A']) v[++tp]=y,w[y]=1;
	}
    }
    for(int i=1;i<=n;i++)
	if(!w[i]) return puts("Yes"),0;
    puts("No");
    return 0;
}
```