#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int n, m;
vector<int> g[200000];
char s[200001];
int deg[200000][2];
bool used[200000];

int main() {
  scanf("%d %d", &n, &m);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    s[i] -= 'A';
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    deg[u][s[v]]++;
    if (u != v) {
      g[v].push_back(u);
      deg[v][s[u]]++;
    }
  }
  queue<int> que;
  for (int i = 0; i < n; i++) {
    if (deg[i][0] == 0 || deg[i][1] == 0) {
      que.push(i);
      used[i] = true;
    }
  }
  while (!que.empty()) {
    n--;
    int u = que.front(); que.pop();
    for (int v : g[u]) if (v != u) {
      if (--deg[v][s[u]] == 0) {
        if (!used[v]) {
          used[v] = true;
          que.push(v);
        }
      }
    }
  }
  puts(n > 0 ? "Yes" : "No");
}

### 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 <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int n, m;
vector<int> g[200000];
char s[200001];
int deg[200000][2];
bool used[200000];

int main() {
  scanf("%d %d", &n, &m);
  scanf("%s", s);
  for (int i = 0; i < n; i++) {
    s[i] -= 'A';
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    deg[u][s[v]]++;
    if (u != v) {
      g[v].push_back(u);
      deg[v][s[u]]++;
    }
  }
  queue<int> que;
  for (int i = 0; i < n; i++) {
    if (deg[i][0] == 0 || deg[i][1] == 0) {
      que.push(i);
      used[i] = true;
    }
  }
  while (!que.empty()) {
    n--;
    int u = que.front(); que.pop();
    for (int v : g[u]) if (v != u) {
      if (--deg[v][s[u]] == 0) {
        if (!used[v]) {
          used[v] = true;
          que.push(v);
        }
      }
    }
  }
  puts(n > 0 ? "Yes" : "No");
}

```