#include <bits/stdc++.h>
using namespace std;
const int maxn = 200200;
int alive[maxn], tot[2][maxn];
vector<int> e[maxn];
int main(){
	int n, m; cin >> n >> m;
	string s; cin >> s;
	for(auto & c : s) c -= 'A';
	for(int i = 0; i < m; i++){
		int x, y; cin >> x >> y; x--; y--;
		e[x].push_back(y);
		e[y].push_back(x);
		tot[s[x]][y]++;
		tot[s[y]][x]++;
	}
	auto getVal = [&](int x){
		int ans = 0;
		if(tot[0][x] > 0) ans += 1;
		if(tot[1][x] > 0) ans += 2;
		return ans;
	};
	set<pair<int,int> > nodes;
	for(int i = 0; i < n; i++){
		alive[i] = 1;
		nodes.insert({getVal(i), i});
	}
	while(!nodes.empty()){
		auto v = *nodes.begin(); nodes.erase(nodes.begin());
		if(v.first == 3) {
			nodes.insert(v);
			break;
		}
		alive[v.second] = 0;
		for(auto nxt : e[v.second]){
			if(alive[nxt]){
				nodes.erase({getVal(nxt), nxt});
				tot[s[v.second]][nxt]--;
				nodes.insert({getVal(nxt), nxt});
			}
		}
	}
	if(nodes.empty()) cout << "No" << endl;
	else cout << "Yes" << endl;
	return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

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


Output

Yes


Input

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


Output

No

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200200;
int alive[maxn], tot[2][maxn];
vector<int> e[maxn];
int main(){
	int n, m; cin >> n >> m;
	string s; cin >> s;
	for(auto & c : s) c -= 'A';
	for(int i = 0; i < m; i++){
		int x, y; cin >> x >> y; x--; y--;
		e[x].push_back(y);
		e[y].push_back(x);
		tot[s[x]][y]++;
		tot[s[y]][x]++;
	}
	auto getVal = [&](int x){
		int ans = 0;
		if(tot[0][x] > 0) ans += 1;
		if(tot[1][x] > 0) ans += 2;
		return ans;
	};
	set<pair<int,int> > nodes;
	for(int i = 0; i < n; i++){
		alive[i] = 1;
		nodes.insert({getVal(i), i});
	}
	while(!nodes.empty()){
		auto v = *nodes.begin(); nodes.erase(nodes.begin());
		if(v.first == 3) {
			nodes.insert(v);
			break;
		}
		alive[v.second] = 0;
		for(auto nxt : e[v.second]){
			if(alive[nxt]){
				nodes.erase({getVal(nxt), nxt});
				tot[s[v.second]][nxt]--;
				nodes.insert({getVal(nxt), nxt});
			}
		}
	}
	if(nodes.empty()) cout << "No" << endl;
	else cout << "Yes" << endl;
	return 0;
}

```