#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,m,d[N][2],tot,head[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
bool vis[N];
struct edge
{
    int u,v;
}e[N];
char s[N];
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&e[i].u,&e[i].v);
        d[e[i].u][s[e[i].v]-'A']++;
        d[e[i].v][s[e[i].u]-'A']++;
        add(e[i].u,e[i].v),add(e[i].v,e[i].u);
    }
    queue<int>q;
    for(int i=1;i<=n;i++)
        if(!(d[i][0]&&d[i][1])) q.push(i),vis[i]=true;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];i;i=nex[i])
        {
            int v=to[i];if(vis[v]) continue;
            d[v][s[u]-'A']--;
            tot--;
            if(!(d[v][0]&&d[v][1])) q.push(v),vis[v]=true;
        }
    }
    int tot=0;
    for(int i=1;i<=n;i++)
        if(!vis[i]) tot++;
    if(tot) printf("Yes\n");
    else printf("No\n");
}

### 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;
const int N=2e5+5;
int n,m,d[N][2],tot,head[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
bool vis[N];
struct edge
{
    int u,v;
}e[N];
char s[N];
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&e[i].u,&e[i].v);
        d[e[i].u][s[e[i].v]-'A']++;
        d[e[i].v][s[e[i].u]-'A']++;
        add(e[i].u,e[i].v),add(e[i].v,e[i].u);
    }
    queue<int>q;
    for(int i=1;i<=n;i++)
        if(!(d[i][0]&&d[i][1])) q.push(i),vis[i]=true;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];i;i=nex[i])
        {
            int v=to[i];if(vis[v]) continue;
            d[v][s[u]-'A']--;
            tot--;
            if(!(d[v][0]&&d[v][1])) q.push(v),vis[v]=true;
        }
    }
    int tot=0;
    for(int i=1;i<=n;i++)
        if(!vis[i]) tot++;
    if(tot) printf("Yes\n");
    else printf("No\n");
}

```