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 <iostream>
#include <vector>
#include <cmath>
#include <string>
#include <iomanip>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <functional>
#include <algorithm>
#include <climits>
#include <sstream>
#include <complex>

using namespace std;

using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
using vvi = vector<vi>;
using iii = pair<int, ii>;
using viii = vector<iii>;

using ll = long long int;
using vc = vector<char>;
using vb = vector<bool>;
using ib = pair<int, bool>;
using vib = vector<ib>;

#define MOD 1000 * 1000 * 1000 + 7
#define INF 1000 * 1000 * 1000
#define EMPTY -10
#define mp(a, b) make_pair(a,b)

bool solve(map<int, pair<set<int>, set<int>>> & A, map<int, pair<set<int>, set<int>>> & B) {
	bool pos = false;
	//Recorremos todas las posiciones de un lado
	for (auto it = A.begin(); it != A.end(); ++it) {
		//Para cada posición buscamos todas aquellas del mismo lado con las que estén conectadas
		for (auto it2 = it->second.first.begin(); it2 != it->second.first.end(); ++it2) {
			for (auto it3 = it->second.second.begin(); it3 != it->second.second.end(); ++it3) {
				for (auto it4 = B[*it3].second.begin(); it4 != B[*it3].second.end(); ++it4) {

					if (A[*it2].second.find(*it4) != A[*it2].second.end()) {
						pos = true;
						goto end;
					}
					else if ( it->second.second.find(*it4) != it->second.second.end() &&
						A[*it2].second.find(*it3) != A[*it2].second.end()) {
						pos = true;
						goto end;
					}
				}
			}
		}
	}
end:;
	return pos;
}

bool resuelveCaso() {
	int n, m;
	cin >> n >> m;
	if (!cin) return false;

	string lact;
	cin >> lact;

	map<int, pair<set<int>, set<int>>> A;
	map<int, pair<set<int>, set<int>>> B;

	for (int i = 0; i < n; ++i) {
		set<int> aux1, aux2;
		if (lact[i] == 'A') A.insert(mp(i + 1,mp(aux1, aux2)));
		else B.insert(mp(i + 1, mp(aux1, aux2)));
	}

	for (int i = 0; i < m; ++i) {
		int ini, fin;
		cin >> ini >> fin;
		--ini; --fin;
		if (ini == fin) {
			if (lact[ini] == 'A') {
				A[ini + 1].first.insert(ini + 1);
			}
			else {
				B[ini + 1].second.insert(ini + 1);
			}
		}
		else {
			if (lact[min(ini, fin)] == 'A') {
				if (lact[max(ini, fin)] == 'A') {
					A[min(ini, fin) + 1].first.insert(max(ini, fin) + 1);
				}
				else {
					A[min(ini, fin) + 1].second.insert(max(ini, fin) + 1);
					B[max(ini, fin) + 1].first.insert(min(ini, fin) + 1);
				}
			}
			else {
				if (lact[max(ini, fin)] == 'A') {
					B[min(ini, fin) + 1].first.insert(max(ini, fin) + 1);
					A[max(ini, fin) + 1].second.insert(min(ini, fin) + 1);
				}
				else {
					B[min(ini, fin) + 1].second.insert(max(ini, fin) + 1);
				}
			}
		}
	}

	if (solve(A, B)) cout << "Yes\n";
	else cout << "No\n";

	return true;
}

int main() {
	while (resuelveCaso()) {}
	return 0;
}