#include <cstdio>
using namespace std;

const int maxn=4e5+5;
int n, m;
char s[maxn];
struct Edge{
    int to, nxt;
}e[maxn*2];
int cnte, fir[maxn], in[maxn];
void addedge(int x, int y){
    Edge &ed=e[++cnte]; in[y]++;
    ed.to=y; ed.nxt=fir[x]; fir[x]=cnte; }
int q[maxn], hd, tl;

int main(){
    scanf("%d%d%s", &n, &m, s+1); int x, y;
    for (int i=1; i<=m; ++i){
        scanf("%d%d", &x, &y);
        if (s[x]==s[y]){  //边为0
            addedge(x*2+1, y*2);
            addedge(y*2+1, x*2); }
        else{
            addedge(x*2, y*2+1);
            addedge(y*2, x*2+1); }
    }
    for (int i=2; i<=n*2+1; ++i)
        if (!in[i]) q[tl++]=i;
    int u, v;
    while (hd<tl){
        u=q[hd++];
        for (int i=fir[u]; i; i=e[i].nxt){
            --in[v=e[i].to]; if (!in[v]) q[tl++]=v; }
    }
    for (int i=2; i<=n*2+1; ++i)
        if (in[i]){ puts("Yes"); return 0; }
    puts("No");
    return 0;
}

### Prompt

In Cpp, your task is to solve 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 <cstdio>
using namespace std;

const int maxn=4e5+5;
int n, m;
char s[maxn];
struct Edge{
    int to, nxt;
}e[maxn*2];
int cnte, fir[maxn], in[maxn];
void addedge(int x, int y){
    Edge &ed=e[++cnte]; in[y]++;
    ed.to=y; ed.nxt=fir[x]; fir[x]=cnte; }
int q[maxn], hd, tl;

int main(){
    scanf("%d%d%s", &n, &m, s+1); int x, y;
    for (int i=1; i<=m; ++i){
        scanf("%d%d", &x, &y);
        if (s[x]==s[y]){  //边为0
            addedge(x*2+1, y*2);
            addedge(y*2+1, x*2); }
        else{
            addedge(x*2, y*2+1);
            addedge(y*2, x*2+1); }
    }
    for (int i=2; i<=n*2+1; ++i)
        if (!in[i]) q[tl++]=i;
    int u, v;
    while (hd<tl){
        u=q[hd++];
        for (int i=fir[u]; i; i=e[i].nxt){
            --in[v=e[i].to]; if (!in[v]) q[tl++]=v; }
    }
    for (int i=2; i<=n*2+1; ++i)
        if (in[i]){ puts("Yes"); return 0; }
    puts("No");
    return 0;
}

```