#include<bits/stdc++.h>
#define maxn 400005

using namespace std;
const int E=400;
char s[maxn];
int a[maxn],b[maxn],n,m,r,rr,d[maxn][2];
bool vis[maxn];
vector<int> h[maxn];

int main(){
    cin >> n >> m;
    scanf("%s",s);
    for (int i=1;i<=n;i++) a[i]=(s[i-1]=='A');
    for (int i=0;i<m;i++){
        int u,v; scanf("%d%d",&u,&v);
        h[u].push_back(v);
        h[v].push_back(u);
        d[u][a[v]]++; d[v][a[u]]++;
    }
    queue <int> q;
    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; rr++;
        for (int i=0;i<h[u].size();i++){
            int v=h[u][i];
            d[v][a[u]]--;
            if (!d[v][a[u]]) q.push(v);
        }
    }
    if (rr<n) puts("Yes"); else puts("No");
    return 0;
}

### 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 maxn 400005

using namespace std;
const int E=400;
char s[maxn];
int a[maxn],b[maxn],n,m,r,rr,d[maxn][2];
bool vis[maxn];
vector<int> h[maxn];

int main(){
    cin >> n >> m;
    scanf("%s",s);
    for (int i=1;i<=n;i++) a[i]=(s[i-1]=='A');
    for (int i=0;i<m;i++){
        int u,v; scanf("%d%d",&u,&v);
        h[u].push_back(v);
        h[v].push_back(u);
        d[u][a[v]]++; d[v][a[u]]++;
    }
    queue <int> q;
    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; rr++;
        for (int i=0;i<h[u].size();i++){
            int v=h[u][i];
            d[v][a[u]]--;
            if (!d[v][a[u]]) q.push(v);
        }
    }
    if (rr<n) puts("Yes"); else puts("No");
    return 0;
}

```