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

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n, m; cin >> n >> m;
    string s; cin >> s;
    vector<int> e[n];
    vector<vector<int>> cnt(n, vector<int>(2, 0));
    vector<bool> dead(n, 0);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        a--, b--;
        e[a].emplace_back(b);
        e[b].emplace_back(a);
        cnt[a][s[b]%2]++, cnt[b][s[a]%2]++;
    }
    queue<int> q;
    for (int i = 0; i < n; i++) {
        if (cnt[i][0] * cnt[i][1] == 0) q.emplace(i), dead[i] = 1;
    }
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        int c = s[curr] % 2;
        for (auto &to: e[curr]) {
            if (dead[to]) continue;
            if (--cnt[to][c] == 0) {
                dead[to] = 1; q.emplace(to);
            }
        }
    }
    bool ok = 0;
    for (int i = 0; i < n; i++) ok |= !dead[i];
    if (ok) cout << "Yes\n";
    else cout << "No\n";
    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>
using namespace std;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n, m; cin >> n >> m;
    string s; cin >> s;
    vector<int> e[n];
    vector<vector<int>> cnt(n, vector<int>(2, 0));
    vector<bool> dead(n, 0);
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        a--, b--;
        e[a].emplace_back(b);
        e[b].emplace_back(a);
        cnt[a][s[b]%2]++, cnt[b][s[a]%2]++;
    }
    queue<int> q;
    for (int i = 0; i < n; i++) {
        if (cnt[i][0] * cnt[i][1] == 0) q.emplace(i), dead[i] = 1;
    }
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        int c = s[curr] % 2;
        for (auto &to: e[curr]) {
            if (dead[to]) continue;
            if (--cnt[to][c] == 0) {
                dead[to] = 1; q.emplace(to);
            }
        }
    }
    bool ok = 0;
    for (int i = 0; i < n; i++) ok |= !dead[i];
    if (ok) cout << "Yes\n";
    else cout << "No\n";
    return 0;
}
```