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

const int N = 201001;

int n, m, a[N][2], b[N];
char s[N];
vector<int> g[N];
queue<int> q;

int main(void) {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for(int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    a[u][s[v] - 'A']++;
    a[v][s[u] - 'A']++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for(int i = 1; i <= n; i++)
    if(!a[i][0] || !a[i][1])
      q.push(i), b[i] = 1;
  while(!q.empty()) {
    int x = q.front(); q.pop();
    for(auto v : g[x])
      if(!b[v] && !--a[v][s[x] - 'A'])
	q.push(v), b[v] = 1;
  }
  for(int i = 1; i <= n; i++)
    if(!b[i]) return puts("Yes"), 0;
  return puts("No"), 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 <bits/stdc++.h>
using namespace std;

const int N = 201001;

int n, m, a[N][2], b[N];
char s[N];
vector<int> g[N];
queue<int> q;

int main(void) {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for(int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    a[u][s[v] - 'A']++;
    a[v][s[u] - 'A']++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for(int i = 1; i <= n; i++)
    if(!a[i][0] || !a[i][1])
      q.push(i), b[i] = 1;
  while(!q.empty()) {
    int x = q.front(); q.pop();
    for(auto v : g[x])
      if(!b[v] && !--a[v][s[x] - 'A'])
	q.push(v), b[v] = 1;
  }
  for(int i = 1; i <= n; i++)
    if(!b[i]) return puts("Yes"), 0;
  return puts("No"), 0;
}

```