#include <bits/stdc++.h>

using namespace std;
#define pb push_back
const int mn=2e5+5;

int n,m;
char s[mn];

int a[mn],b[mn];
bool vis[mn];
vector<int> g[mn];
queue<int> q;

int main() {
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1; i<=m; i++) {
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].pb(v);
        g[v].pb(u);
        if (s[u]=='A') a[v]++;
        else b[v]++;
        if (s[v]=='A') a[u]++;
        else b[u]++;
    }

    for(int i=1; i<=n; i++) {
        if (a[i]==0||b[i]==0) q.push(i),vis[i]=1;
    }
    while(!q.empty()) {
        int u=q.front();
        q.pop();
        for(auto &i:g[u]) {
            if (s[u]=='A') {
                a[i]--;
                if (!vis[i]&&a[i]==0) q.push(i),vis[i]=1;
            }
            else {
                b[i]--;
                if (!vis[i]&&b[i]==0) q.push(i),vis[i]=1;
            }
        }
    }
    int ans=0;
    for(int i=1; i<=n; i++)
        if (!vis[i]) ans=1;
    if (ans) puts("Yes");
    else puts("No");

    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
#define pb push_back
const int mn=2e5+5;

int n,m;
char s[mn];

int a[mn],b[mn];
bool vis[mn];
vector<int> g[mn];
queue<int> q;

int main() {
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1; i<=m; i++) {
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].pb(v);
        g[v].pb(u);
        if (s[u]=='A') a[v]++;
        else b[v]++;
        if (s[v]=='A') a[u]++;
        else b[u]++;
    }

    for(int i=1; i<=n; i++) {
        if (a[i]==0||b[i]==0) q.push(i),vis[i]=1;
    }
    while(!q.empty()) {
        int u=q.front();
        q.pop();
        for(auto &i:g[u]) {
            if (s[u]=='A') {
                a[i]--;
                if (!vis[i]&&a[i]==0) q.push(i),vis[i]=1;
            }
            else {
                b[i]--;
                if (!vis[i]&&b[i]==0) q.push(i),vis[i]=1;
            }
        }
    }
    int ans=0;
    for(int i=1; i<=n; i++)
        if (!vis[i]) ans=1;
    if (ans) puts("Yes");
    else puts("No");

    return 0;
}

```