#include <bits/stdc++.h>
using namespace std;

int N, M;
string S;
const int MAX = 200000;
vector<int> edges[MAX];
int numA[MAX], numB[MAX];
bool deleted[MAX];

void del(int i){
    if(deleted[i]) return;
    deleted[i] = true;
    for(auto j : edges[i]){
        if(deleted[j]) continue;
        (S[i] == 'A' ? numA[j] : numB[j]) -= 1;
        if(numA[j] == 0 || numB[j] == 0) del(j);
    }
}

int main(){
    int i, j, k;
    cin >> N >> M >> S;
    for(i=0; i<M; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
        (S[b] == 'A' ? numA[a] : numB[a]) += 1;
        (S[a] == 'A' ? numA[b] : numB[b]) += 1;
    }

    for(i=0; i<N; i++){
        if(numA[i] == 0 || numB[i] == 0) del(i);
    }
    for(i=0; i<N; i++){
        if(!deleted[i]){
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
string S;
const int MAX = 200000;
vector<int> edges[MAX];
int numA[MAX], numB[MAX];
bool deleted[MAX];

void del(int i){
    if(deleted[i]) return;
    deleted[i] = true;
    for(auto j : edges[i]){
        if(deleted[j]) continue;
        (S[i] == 'A' ? numA[j] : numB[j]) -= 1;
        if(numA[j] == 0 || numB[j] == 0) del(j);
    }
}

int main(){
    int i, j, k;
    cin >> N >> M >> S;
    for(i=0; i<M; i++){
        int a, b;
        cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
        (S[b] == 'A' ? numA[a] : numB[a]) += 1;
        (S[a] == 'A' ? numA[b] : numB[b]) += 1;
    }

    for(i=0; i<N; i++){
        if(numA[i] == 0 || numB[i] == 0) del(i);
    }
    for(i=0; i<N; i++){
        if(!deleted[i]){
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
    return 0;
}
```