#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair

typedef long long ll;
typedef pair<int, int> ii;
typedef pair<ll, ll> l4;

const int N = 2e5+1;
const int NN = 2 * N;
char s[N+1];
int n, m;
vector<int> g[NN];
int in[NN];
inline void add(int x, int y)
{
  g[x].pb(y);
  ++in[y];
}
int main()
{
  scanf("%d %d %s", &n, &m, s+1);
  for (int i = 1; i <= m; ++i)
    {
      int a, b; scanf("%d %d", &a, &b);
      if (s[a] != s[b])
        add(2*a-1, 2*b), add(2*b-1, 2*a);
      else
        add(2*a, 2*b-1), add(2*b, 2*a-1);
    }
  queue<int> q;
  for (int i = 1; i <= 2*n; ++i)
    if (in[i] == 0) q.push(i);
  int vis = 0;
  while (!q.empty())
    {
      ++vis;
      int cur = q.front(); q.pop();
      for (auto nxt : g[cur])
        if (--in[nxt] == 0)
          q.push(nxt);
    }
  puts(vis == 2*n?"No":"Yes");

}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
#define pb push_back
#define mp make_pair

typedef long long ll;
typedef pair<int, int> ii;
typedef pair<ll, ll> l4;

const int N = 2e5+1;
const int NN = 2 * N;
char s[N+1];
int n, m;
vector<int> g[NN];
int in[NN];
inline void add(int x, int y)
{
  g[x].pb(y);
  ++in[y];
}
int main()
{
  scanf("%d %d %s", &n, &m, s+1);
  for (int i = 1; i <= m; ++i)
    {
      int a, b; scanf("%d %d", &a, &b);
      if (s[a] != s[b])
        add(2*a-1, 2*b), add(2*b-1, 2*a);
      else
        add(2*a, 2*b-1), add(2*b, 2*a-1);
    }
  queue<int> q;
  for (int i = 1; i <= 2*n; ++i)
    if (in[i] == 0) q.push(i);
  int vis = 0;
  while (!q.empty())
    {
      ++vis;
      int cur = q.front(); q.pop();
      for (auto nxt : g[cur])
        if (--in[nxt] == 0)
          q.push(nxt);
    }
  puts(vis == 2*n?"No":"Yes");

}

```