#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
char str[400005];
int a[400005],b[400005],d[400005][2];
int n,m,ans;
bool vis[400005];
vector<int>V[400005];
queue<int>Q;
int main()
{
	scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    for (int i=1;i<=n;i++) a[i]=(str[i]=='A');
    for (int i=1,x,y;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        V[x].push_back(y);
        V[y].push_back(x);
        d[x][a[y]]++;
        d[y][a[x]]++;
    }
    for (int i=1;i<=n;i++)
    	if (!d[i][0] || !d[i][1]) Q.push(i);
    while (!Q.empty())
    {
        int u=Q.front();Q.pop();
        if (vis[u]) continue;
        vis[u]=true;
        ans++;
        for (int i=0;i<V[u].size();i++)
        {
            int v=V[u][i];
            d[v][a[u]]--;
            if (!d[v][a[u]]) Q.push(v);
        }
    }
    if (ans<n) printf("Yes\n"); else printf("No\n");
    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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
char str[400005];
int a[400005],b[400005],d[400005][2];
int n,m,ans;
bool vis[400005];
vector<int>V[400005];
queue<int>Q;
int main()
{
	scanf("%d%d",&n,&m);
    scanf("%s",str+1);
    for (int i=1;i<=n;i++) a[i]=(str[i]=='A');
    for (int i=1,x,y;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        V[x].push_back(y);
        V[y].push_back(x);
        d[x][a[y]]++;
        d[y][a[x]]++;
    }
    for (int i=1;i<=n;i++)
    	if (!d[i][0] || !d[i][1]) Q.push(i);
    while (!Q.empty())
    {
        int u=Q.front();Q.pop();
        if (vis[u]) continue;
        vis[u]=true;
        ans++;
        for (int i=0;i<V[u].size();i++)
        {
            int v=V[u][i];
            d[v][a[u]]--;
            if (!d[v][a[u]]) Q.push(v);
        }
    }
    if (ans<n) printf("Yes\n"); else printf("No\n");
    return 0;
}

```