#include<bits/stdc++.h>
using namespace std;
const int nn=400005;
int n,m,tot=0,f[nn],col[nn]; char co[nn]; ;
struct dat{
    int from,to,next;
}e[nn*2];
void adde(int x,int y){ e[++tot]=(dat){x,y,f[x]}; f[x]=tot; }
bool dfs(int now){
    col[now]=1;
    for(int i=f[now];i;i=e[i].next){
        int to=e[i].to;
        if(col[to]==1){ return 1;}
        if(col[to]==0&&dfs(to)) return 1;
    }
    col[now]=2;
    return 0;
}
int main(){
    scanf("%d%d",&n,&m); scanf("%s",co+1);
    for(int a,b,i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        if(co[a]==co[b]) adde(a,b+n),adde(b,a+n);
        if(co[a]!=co[b]) adde(a+n,b),adde(b+n,a);
    }
    for(int i=1;i<=n;i++)if(!col[i])if(dfs(i)){printf("Yes\n"); return 0;}
    printf("No\n");
    return 0;
}
### Prompt

Please formulate 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 nn=400005;
int n,m,tot=0,f[nn],col[nn]; char co[nn]; ;
struct dat{
    int from,to,next;
}e[nn*2];
void adde(int x,int y){ e[++tot]=(dat){x,y,f[x]}; f[x]=tot; }
bool dfs(int now){
    col[now]=1;
    for(int i=f[now];i;i=e[i].next){
        int to=e[i].to;
        if(col[to]==1){ return 1;}
        if(col[to]==0&&dfs(to)) return 1;
    }
    col[now]=2;
    return 0;
}
int main(){
    scanf("%d%d",&n,&m); scanf("%s",co+1);
    for(int a,b,i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        if(co[a]==co[b]) adde(a,b+n),adde(b,a+n);
        if(co[a]!=co[b]) adde(a+n,b),adde(b+n,a);
    }
    for(int i=1;i<=n;i++)if(!col[i])if(dfs(i)){printf("Yes\n"); return 0;}
    printf("No\n");
    return 0;
}
```