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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <queue>
#include <stack>
#include <cstdlib>
#include <map>
#include <iomanip>
#include <set>
#include <stdio.h>
#include <ctype.h>
#include <random>
#include <string.h>
#include <cmath>
#include <unordered_map>
#include <cstdio>
using namespace std;
#define all(vec) vec.begin(),vec.end()
typedef long long ll;

ll gcd(ll x, ll y) {
	if (y == 0)return x;
	return gcd(y, x%y);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}
ll kai(ll x, ll y, ll m) {
	ll res = 1;
	for (ll i = x - y + 1; i <= x; i++) {
		res *= i; res %= m;
	}
	return res;
}
ll mod_pow(ll x, ll y, ll m) {
	ll res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}

ll comb(ll x, ll y, ll m) {
	if (y > x)return 0;
	return kai(x, y, m) * mod_pow(kai(y, y, m), m - 2, m) % m;
}

int n, m, a[200010], b[200010];
int d[400010];
bool bl[400010];
vector<int> vec[400010];
string s;
bool ans;

void dfs(int v, int k) {
	d[v] = k;
	bl[v] = true;
	int p = k + 1;
	if (p == 5)p -= 4;
	for (int i = 0; i < (int)vec[v].size(); i++) {
		if (d[vec[v][i]] == p) {
			ans = true;
			return;
		}
		if (!d[vec[v][i]] && ((p <= 2 && s[vec[v][i] / 2] == 'A') || (3 <= p && s[vec[v][i] / 2] == 'B')))dfs(vec[v][i], p);
	}
	d[v] = 0;
}

signed main() {
	cin >> n >> m >> s;
	for (int i = 0; i < m; i++) {
		cin >> a[i] >> b[i]; a[i]--; b[i]--; a[i] *= 2; b[i] *= 2;
		if (a[i] == b[i]) {
			vec[a[i]].push_back(a[i] + 1);
			vec[a[i] + 1].push_back(a[i]);
			continue;
		}
		vec[a[i]].push_back(b[i]);
		vec[a[i] + 1].push_back(b[i]);
		vec[a[i]].push_back(b[i] + 1);
		vec[a[i] + 1].push_back(b[i] + 1);
		vec[b[i]].push_back(a[i]);
		vec[b[i] + 1].push_back(a[i]);
		vec[b[i]].push_back(a[i] + 1);
		vec[b[i] + 1].push_back(a[i] + 1);
	}
	for (int i = 0; i < 2*n; i++) {
		if (bl[i])continue;
		if (s[i / 2] == 'A')dfs(i, 1);
		else dfs(i, 3);
	}
	if (ans)cout << "Yes" << endl;
	else cout << "No" << endl;
}