#include <bits/stdc++.h>
#define LL long long
#define MEM(x,y) memset(x,y,sizeof(x))
#define MOD(x) ((x)%1000000007)

using namespace std;
const int maxn = 2e5 + 7;
char s[maxn];
int deg[maxn][2], now, del[maxn];
set<int>E[maxn];

void dfs(int x)
{
    if(del[x] || (deg[x][0] > 0 && deg[x][1] > 0)) return;
    now--; del[x] = 1;
    //cout << "del " << x << endl;
    for(int tmp : E[x])
    {
        deg[tmp][s[x - 1] - 'A']--;
        deg[x][s[tmp - 1] - 'A']--;
        dfs(tmp);
    }
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int n, m;
    cin >> n >> m;
    cin >> s;
    while(m--)
    {
        int u, v;
        cin >> u >> v;
        if(!E[u].count(v))
        {
            deg[u][s[v - 1] - 'A']++; E[u].insert(v);
        }
        if(!E[v].count(u))
        {
            deg[v][s[u - 1] - 'A']++; E[v].insert(u);
        }
    }
    now = n;
    for(int i = 1; i <= n; i++) dfs(i);
    if(now > 0) puts("Yes");
    else puts("No");
    return 0;
}

### 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 <bits/stdc++.h>
#define LL long long
#define MEM(x,y) memset(x,y,sizeof(x))
#define MOD(x) ((x)%1000000007)

using namespace std;
const int maxn = 2e5 + 7;
char s[maxn];
int deg[maxn][2], now, del[maxn];
set<int>E[maxn];

void dfs(int x)
{
    if(del[x] || (deg[x][0] > 0 && deg[x][1] > 0)) return;
    now--; del[x] = 1;
    //cout << "del " << x << endl;
    for(int tmp : E[x])
    {
        deg[tmp][s[x - 1] - 'A']--;
        deg[x][s[tmp - 1] - 'A']--;
        dfs(tmp);
    }
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int n, m;
    cin >> n >> m;
    cin >> s;
    while(m--)
    {
        int u, v;
        cin >> u >> v;
        if(!E[u].count(v))
        {
            deg[u][s[v - 1] - 'A']++; E[u].insert(v);
        }
        if(!E[v].count(u))
        {
            deg[v][s[u - 1] - 'A']++; E[v].insert(u);
        }
    }
    now = n;
    for(int i = 1; i <= n; i++) dfs(i);
    if(now > 0) puts("Yes");
    else puts("No");
    return 0;
}

```