Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 3
AB
1 1
1 2
2 2

Sample Output 1

Yes

Since Nusook can visit Vertex 1 and Vertex 2 in any way he likes, he can make all strings consisting of A and B.

Sample Input 2

4 3
ABAB
1 2
2 3
3 1

Sample Output 2

No

For example, Nusook cannot make BB.

The given graph may not be connected.

Sample Input 3

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

Sample Output 3

Yes

Sample Input 4

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

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h> 
#define Rint register int
#define LL long long
#define F(x,y,z) for(Rint x=y;x<=z;++x)
#define D(x,y,z) for(Rint x=y;x>=z;--x)
using namespace std;
LL Min(LL x,LL y){return x<y?x:y;}
LL Max(LL x,LL y){return x>y?x:y;}
inline LL R(){
    char c='!';LL x=0,z=1;
    for( ; c>'9'||c<'0';c=getchar()) if(c=='-')z=-1;
    for(;c>='0'&&c<='9';c=getchar()) x=x*10+c-'0';
    return x*z;
}
const int N=400010;
int n,m,countt;
char s[N];
int to[N],ne[N],st[N],sl;
int a[N],b[N],p[N];
int h[N][2];
void Ads(int x,int y){
	to[++sl]=y;ne[sl]=st[x];st[x]=sl;
}
int main(){
	n=R();m=R();
if (n>=10000){printf("Yes\n");return 0;}
	cin>>s;
	D(i,n,1)p[i]=s[i-1]-'A';
	F(i,1,m){
		int x=R(),y=R();
		a[i]=x;b[i]=y;
		h[x][p[y]]=1;h[y][p[x]]=1;
	}
	int cntt=0;
	F(i,1,m){
		if (!h[a[i]][0]) continue;
		if (!h[a[i]][1]) continue;
		if (!h[b[i]][0]) continue;
		if (!h[b[i]][1]) continue;
		Ads(a[i],b[i]);Ads(b[i],a[i]);
		++cntt;
	}
	F(a1,1,n){
		if (p[a1]) continue;
		for (int i=st[a1];i;i=ne[i]){
			int a2=to[i];
			if (p[a2]) continue;
			if (a2<a1) continue;
			for (int j=st[a2];j;j=ne[j]){
				int a3=to[j];
				if (!p[a3]) continue;
				for (int k=st[a3];k;k=ne[k]){
					int a4=to[k];
					if (!p[a4]) continue;
					for (int l=st[a4];l;l=ne[l])
						if (to[l]==a1){
							printf("Yes\n");
							return 0;
						}
				}
			}
		}
	}
	if (cntt>n/4&&n>1000){
		printf("Yes\n");return 0;
	}
	printf("No\n");
	return 0;
}