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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class Main {
	public static class Node {
		public int id;
		public char c;
		public Set<Integer> dsts = new HashSet<>();

		public Node(int id, char c) {
			this.id = id;
			this.c = c;
		}
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String[] tokens = in.nextLine().split(" ");
		int N = Integer.parseInt(tokens[0]);
		int M = Integer.parseInt(tokens[1]);
		String s = in.nextLine();
		char[] nodeType = new char[N];
		for (int i = 0; i < N; ++i) {
			nodeType[i] = s.charAt(i);
		}
		// Nodeを作成する
		Map<Integer, Node> map = new HashMap<>();
		for (int i = 1; i <= N; ++i) {
			Node node = new Node(i, nodeType[i - 1]);
			map.put(i, node);
		}
		// Edgeを読み込む
		for (int i = 0; i < M; ++i) {
			tokens = in.nextLine().split(" ");
			int a = Integer.parseInt(tokens[0]);
			int b = Integer.parseInt(tokens[1]);
			map.get(a).dsts.add(b);
			map.get(b).dsts.add(a);
		}

		// edgeの本数が1以下のノードを全部削除する
		while (true) {
			// 隣接するノードの種類が1種類以下のものを削除する
			Set<Integer> removeKeys = new HashSet<>();
			for (Integer key : map.keySet()) {
				Node node = map.get(key);
				Set<Character> types = new HashSet<>();
				for (Integer next : node.dsts) {
					types.add(map.get(next).c);
				}
				if (types.size() <= 1) {
					removeKeys.add(key);
				}
			}
			if (removeKeys.isEmpty()) {
				break;
			}
			for (Integer key : removeKeys) {
				map.remove(key);
			}
			for (Integer key : map.keySet()) {
				map.get(key).dsts.removeAll(removeKeys);
			}
		}
		if (map.isEmpty()) {
			System.out.println("No");
		} else {
			System.out.println("Yes");
		}
		in.close();
	}
}