#include <bits/stdc++.h>

using namespace std;

int N, M, cnt[200005][2], vis[200005];
char ch[200005];
vector<int> G[200005];
queue<int> que;

int main() {
  scanf("%d%d", &N, &M);
  scanf("%s", ch + 1);
  for (int i = 1, u, v; i <= M; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v), cnt[u][ch[v] - 'A']++;
    G[v].push_back(u), cnt[v][ch[u] - 'A']++;
  }
  for (int i = 1; i <= N; i++)
    if (!cnt[i][0] || !cnt[i][1]) vis[i] = true, que.push(i);
  while (!que.empty()) {
    int now = que.front();
    que.pop();
    for (auto to : G[now]) {
      if (vis[to]) continue;
      cnt[to][ch[now] - 'A']--;
      if (!cnt[to][0] || !cnt[to][1]) vis[to] = true, que.push(to);
    }
  }
  bool flag = false;
  for (int i = 1; i <= N && !flag; i++)
    if (!vis[i]) flag = true;
  puts(flag ? "Yes" : "No");
  return 0;
}

### Prompt

Create a solution in CPP for 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 N, M, cnt[200005][2], vis[200005];
char ch[200005];
vector<int> G[200005];
queue<int> que;

int main() {
  scanf("%d%d", &N, &M);
  scanf("%s", ch + 1);
  for (int i = 1, u, v; i <= M; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v), cnt[u][ch[v] - 'A']++;
    G[v].push_back(u), cnt[v][ch[u] - 'A']++;
  }
  for (int i = 1; i <= N; i++)
    if (!cnt[i][0] || !cnt[i][1]) vis[i] = true, que.push(i);
  while (!que.empty()) {
    int now = que.front();
    que.pop();
    for (auto to : G[now]) {
      if (vis[to]) continue;
      cnt[to][ch[now] - 'A']--;
      if (!cnt[to][0] || !cnt[to][1]) vis[to] = true, que.push(to);
    }
  }
  bool flag = false;
  for (int i = 1; i <= N && !flag; i++)
    if (!vis[i]) flag = true;
  puts(flag ? "Yes" : "No");
  return 0;
}

```