#include <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <functional>
using namespace std;

int main() {
    int n, m; cin >> n >> m;
    string s; cin >> s;
    vector<vector<int>> g(n);
    vector<int> cntA(n), cntB(n);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b; a--, b--;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
        (s[b] == 'A' ? cntA[a] : cntB[a])++;
        (s[a] == 'A' ? cntA[b] : cntB[b])++;
    }
    vector<int> alive(n, true);
    int cnt = n;
    function<void(int)> del = [&](int u) {
        if (!alive[u]) return ;
        alive[u] = false, cnt--;
        for (int v: g[u]) {
            (s[u] == 'A' ? cntA[v] : cntB[v])--;
            if (alive[v] && (cntA[v] == 0 || cntB[v] == 0)) del(v);
        }
    };
    for (int i = 0; i < n; i++) if (cntA[i] == 0 || cntB[i] == 0) del(i);
    cout << (cnt ? "Yes" : "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 <algorithm>
#include <iostream>
#include <vector>
#include <set>
#include <functional>
using namespace std;

int main() {
    int n, m; cin >> n >> m;
    string s; cin >> s;
    vector<vector<int>> g(n);
    vector<int> cntA(n), cntB(n);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b; a--, b--;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
        (s[b] == 'A' ? cntA[a] : cntB[a])++;
        (s[a] == 'A' ? cntA[b] : cntB[b])++;
    }
    vector<int> alive(n, true);
    int cnt = n;
    function<void(int)> del = [&](int u) {
        if (!alive[u]) return ;
        alive[u] = false, cnt--;
        for (int v: g[u]) {
            (s[u] == 'A' ? cntA[v] : cntB[v])--;
            if (alive[v] && (cntA[v] == 0 || cntB[v] == 0)) del(v);
        }
    };
    for (int i = 0; i < n; i++) if (cntA[i] == 0 || cntB[i] == 0) del(i);
    cout << (cnt ? "Yes" : "No") << endl;
    return 0;
}

```