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 <bits/stdc++.h>
// #include "lib.h"
using namespace std;
#define LLINF 1234567890987654321
#define INF 1234567890
#define pb push_back
#define ins insert
#define f first
#define s second
#define db 0
#define MAXN 200006
#define MAXK 200006
#define MAXX 200006
#define rep(kk, l1, l2)for(ll kk = l1; kk < l2; kk++)
#define bg(ms) *ms.begin()
#define ed(ms) *prev(ms.end(), 1)
#define addedge(a, b, c) v[a].pb(pi(b, c)); v[b].pb(pi(a, c))
#define ph push
#define all(x) x.begin(), x.end()
typedef long long ll;
typedef pair <ll, ll> pi;
typedef pair <ll, pi> spi;
typedef pair <pi, pi> dpi;
ll n, m, AB[MAXN];  //AB, 0 for A, 1 for B;
string ss;
ll vis[MAXN], can[MAXN];
vector <ll> v[MAXN];
bool dfs(ll x, ll p, ll a, ll b)
{
	vis[x] = 1;
  can[x] = 1;
	for(auto i : v[x])
	{
		if(vis[i] && can[i])
		{
      // cout << x << ' ' << i << '\n';
			if(AB[i] == 0)a++;
      else b++;
      continue;
		}
    if(dfs(i, x, 0, 0))
    {
      a += ((AB[i]==0)?1:0);
      b += ((AB[i]==1)?1:0);
    }
	}
  // cout << x << ' ' << a << ' ' << b << '\n';
  if(a >= 1 && b >= 1)can[x] = 1;
  else can[x] = 0;
  return can[x];
}
int main()
{
	cin >> n >> m >> ss;
	for(ll i = 0; i < ss.length(); i++)AB[i+1] = ((ss[i]=='A') ? 0 : 1);
	for(ll i = 0; i < m; i++)
	{
		ll n1, n2;
		cin >> n1 >> n2;
    if(n1==n2)
    {
      v[n1].pb(n2);
      continue;
    }
		v[n1].pb(n2);
		v[n2].pb(n1);
	}
	for(ll i = 1; i <= n; i++)
	{
		if(vis[i])continue;
    // cout << "hi\n";
		if(dfs(i, -1, 0, 0 ))
		{
			cout << "Yes\n";
			return 0;
		}
	}
	cout << "No\n";
	return 0;
}