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: Accepted)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

const int N=200005;

int n,m,cnt,last[N],a[N];
bool vis[N][2],ins[N][2];
struct edge{int to,next,w;}e[N*2];
char s[N];

void addedge(int u,int v,int w)
{
	e[++cnt].to=v;e[cnt].w=w;e[cnt].next=last[u];last[u]=cnt;
	e[++cnt].to=u;e[cnt].w=w;e[cnt].next=last[v];last[v]=cnt;
}

void dfs(int x,int y)
{
	vis[x][y]=ins[x][y]=1;
	for (int i=last[x];i;i=e[i].next)
		if (e[i].w!=y)
		{
			if (!vis[e[i].to][y^1]) dfs(e[i].to,y^1);
			else if (ins[e[i].to][y^1]) {puts("Yes");exit(0);}
		}
	ins[x][y]=0;
}

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
		if (s[i]=='A') a[i]=0;
		else a[i]=1;
	for (int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		addedge(x,y,a[x]^a[y]);
	}
	for (int i=1;i<=n;i++)
	{
		if (!vis[i][0]) dfs(i,0);
		if (!vis[i][1]) dfs(i,1);
	}
	puts("No");
	return 0;
}