#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

const int N = 200005;

int n, m, u, v, cnt[N][2];
bool chk[N];
char s[N];
vector<int> ve, adj[N];

int main()
{
    scanf("%d %d\n%s\n", &n, &m, s + 1);
    for (int i = 1; i <= n; i++)
        s[i] -= 'A';
    for (int i = 1; i <= m; i++)
    {
        scanf("%d %d\n", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
        cnt[u][s[v]]++;
        cnt[v][s[u]]++;
    }
    for (int i = 1; i <= n; i++)
        if (cnt[i][0] == 0 || cnt[i][1] == 0)
        {
            ve.push_back(i);
            chk[i] = true;
        }
    for (unsigned int i = 0; i < ve.size(); i++)
    {
        int u = ve[i];
        for (int &v : adj[u])
            if (!chk[v] && (--cnt[v][s[u]]) == 0)
            {
                ve.push_back(v);
                chk[v] = true;
            }
    }
    printf("%s\n", ve.size() == 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 <iostream>
#include <cstdio>
#include <vector>
using namespace std;

const int N = 200005;

int n, m, u, v, cnt[N][2];
bool chk[N];
char s[N];
vector<int> ve, adj[N];

int main()
{
    scanf("%d %d\n%s\n", &n, &m, s + 1);
    for (int i = 1; i <= n; i++)
        s[i] -= 'A';
    for (int i = 1; i <= m; i++)
    {
        scanf("%d %d\n", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
        cnt[u][s[v]]++;
        cnt[v][s[u]]++;
    }
    for (int i = 1; i <= n; i++)
        if (cnt[i][0] == 0 || cnt[i][1] == 0)
        {
            ve.push_back(i);
            chk[i] = true;
        }
    for (unsigned int i = 0; i < ve.size(); i++)
    {
        int u = ve[i];
        for (int &v : adj[u])
            if (!chk[v] && (--cnt[v][s[u]]) == 0)
            {
                ve.push_back(v);
                chk[v] = true;
            }
    }
    printf("%s\n", ve.size() == n ? "No" : "Yes");
}
```