#include <bits/stdc++.h>
using namespace std;
int n,m,a,b,g[200002][2],is_del[200002];
vector<int> v[200002],u;
string s;

int main (){
    cin >> n >> m;
    cin >> s;
    for (int i = 0; i < m ; i++ ){
        cin >> a >> b ;
        v[a].push_back(b);
        v[b].push_back(a);
        g[a][s[b - 1] == 'A' ] ++;
        g[b][s[a - 1] == 'A' ] ++;
    }
    for(int i=1;i<=n;i++) {
        if(!g[i][0] || !g[i][1]){
            is_del[i] = 1;
            u.push_back(i);
        }
    }
    for (int i=0;i<u.size();i++){
        for(auto j:v[u[i]]){
            g[j][ s[ u[i] - 1 ] == 'A' ] -- ;
            if(!(g[j][ s[ u[i] - 1 ] == 'A' ]) && !is_del[j]){
                u.push_back(j);
                is_del[j] = 1;
            }
        }
    }
    if (u.size() != n){
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}

### 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;
int n,m,a,b,g[200002][2],is_del[200002];
vector<int> v[200002],u;
string s;

int main (){
    cin >> n >> m;
    cin >> s;
    for (int i = 0; i < m ; i++ ){
        cin >> a >> b ;
        v[a].push_back(b);
        v[b].push_back(a);
        g[a][s[b - 1] == 'A' ] ++;
        g[b][s[a - 1] == 'A' ] ++;
    }
    for(int i=1;i<=n;i++) {
        if(!g[i][0] || !g[i][1]){
            is_del[i] = 1;
            u.push_back(i);
        }
    }
    for (int i=0;i<u.size();i++){
        for(auto j:v[u[i]]){
            g[j][ s[ u[i] - 1 ] == 'A' ] -- ;
            if(!(g[j][ s[ u[i] - 1 ] == 'A' ]) && !is_del[j]){
                u.push_back(j);
                is_del[j] = 1;
            }
        }
    }
    if (u.size() != n){
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}

```