#include <bits/stdc++.h>
using namespace std;
string s;
vector<vector<int>> grh, cnt;
vector<int> ckd;
void dfs (int crr) {
  if (ckd[crr]) return;
  if (cnt[crr][0] && cnt[crr][1]) return;
  ckd[crr] = true;
  for (int nxt : grh[crr]) {
    if (ckd[nxt]) continue;
    if (s[crr] == s[nxt]) {
      cnt[crr][0]--;
      cnt[nxt][0]--;
    } else {
      cnt[crr][1]--;
      cnt[nxt][1]--;
    }
    dfs(nxt);
  }
}
int main() {
  int n, m;
  cin >> n >> m >> s;
  grh.resize(n);
  cnt.resize(n, vector<int>(2));
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    grh[a].push_back(b);
    grh[b].push_back(a);
    if (s[a] == s[b]) {
      cnt[a][0]++;
      cnt[b][0]++;
    } else {
      cnt[a][1]++;
      cnt[b][1]++;
    }
  }
  ckd.resize(n, false);
  for (int i = 0; i < n; i++) {
    dfs(i);
  }
  int flg = false;
  for (int i = 0; i < n; i++) {
    if (!ckd[i]) flg = true;
  }
  if (flg) {
    cout << "Yes" << '\n';
  } else {
    cout << "No" << '\n';
  }
  return 0;
}
### Prompt

Your task is to create 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;
string s;
vector<vector<int>> grh, cnt;
vector<int> ckd;
void dfs (int crr) {
  if (ckd[crr]) return;
  if (cnt[crr][0] && cnt[crr][1]) return;
  ckd[crr] = true;
  for (int nxt : grh[crr]) {
    if (ckd[nxt]) continue;
    if (s[crr] == s[nxt]) {
      cnt[crr][0]--;
      cnt[nxt][0]--;
    } else {
      cnt[crr][1]--;
      cnt[nxt][1]--;
    }
    dfs(nxt);
  }
}
int main() {
  int n, m;
  cin >> n >> m >> s;
  grh.resize(n);
  cnt.resize(n, vector<int>(2));
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    grh[a].push_back(b);
    grh[b].push_back(a);
    if (s[a] == s[b]) {
      cnt[a][0]++;
      cnt[b][0]++;
    } else {
      cnt[a][1]++;
      cnt[b][1]++;
    }
  }
  ckd.resize(n, false);
  for (int i = 0; i < n; i++) {
    dfs(i);
  }
  int flg = false;
  for (int i = 0; i < n; i++) {
    if (!ckd[i]) flg = true;
  }
  if (flg) {
    cout << "Yes" << '\n';
  } else {
    cout << "No" << '\n';
  }
  return 0;
}
```