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>
#define ALL(v) v.begin(), v.end()
#define MAX 510000
#define rrep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define rep(i, n) for(ll i = 1; i <= (ll)(n); i++)
#define dcout cout<<fixed<<setprecision(15);
#define mp make_pair
#define pb push_back

using namespace std;
typedef long long int  ll;
typedef pair<ll, ll> P;

constexpr int MOD = 1e9 + 7;
constexpr ll inf = 1LL << 60;

template< typename S, typename T >
inline void chmax(S &a, const T &b) { if(a < b) a = b; }
 
template< typename S, typename T >
inline void chmin(S &a, const T &b) { if(a > b) a = b; }
 

//lcm//
ll gcd(ll x, ll y) {
	if (x == 0) return y;
	return gcd(y%x, x);
}
ll lcm(ll x, ll y) { return x * y / gcd(x, y); }

//a^n mod p//
ll modpow(ll a, ll n, ll p) {
	if(n==0) return (ll)1;
  if (n == 1) return a % p;
  if (n % 2 == 1) return (a * modpow(a, n - 1, p)) % p;
  ll t = modpow(a, n / 2, p);
  return (t * t) % p;
}

//inversemod//
ll modinv(ll a, ll m) {
	if(m==0)return (ll)1;
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}

//Cmonp//
ll fac[MAX], finv[MAX], inv[MAX];
// 
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 
ll COM(ll n, ll k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//LARGE n
ll com(ll n,ll m){
    if(n<m || n<=0 ||m<0){
		return 0;
	}
	if( m==0 || n==m){
		return 1;
	}
	ll k=1;
	for(ll i=1;i<=m;i++){
       k*=(n-i+1); 
	   k%=MOD;
	   k*=modinv(i,MOD);
	   k%=MOD;
	}
	return k;
}

//radP
ll rad(ll u, ll p){
ll cnt=0;
	while(u%p==0){
		u/=p;
		cnt++;
	}
	return cnt;
}


////////////////////////////////////////////////////////////////////

ll n,m;
vector<ll> vec[210000];
string s;
vector<P> cnt;
queue<ll> que{};

void bfs(){
while(que.size()){
	int v=que.front();
	que.pop();
	int att;
	if(s[v]=='A')att=1;
	else att=2;
for(int i=0;i<vec[v].size();i++){
ll vv=vec[v][i];
if(att==1 && cnt[vv].first==1){
	que.push(vv);
}
if(att==2 && cnt[vv].second==1){
	que.push(vv);
}
if(att==1&&cnt[vv].first>=1)cnt[vv].first--;
if(att==2&&cnt[vv].second>=1)cnt[vv].second--;
}


}
return ;
}

int main() {
cin>>n>>m;
cin>>s;

rep(i,m){
	ll a,b;
	cin>>a>>b;
vec[a].pb(b);
if(a!=b)
vec[b].pb(a);
}

cnt.pb(mp(0,0));
rep(i,n){
	ll Ac=0;
	ll Bc=0;
for(int j=0;j<vec[i].size();j++){
	if(s[vec[i][j]-1]=='A'){
Ac++;
	}
	else Bc++;
}
cnt.pb(mp(Ac,Bc));
}

rep(i,n){
	int p=cnt[i].first;
	int q=cnt[i].second;
	if(p==0 || q==0){
		que.push(i);
	}
}

bfs();

ll ans=0;

rep(i,n){
	int p=cnt[i].first;
	int q=cnt[i].second;
	if(p!=0 && q!=0){
		ans++;
	}
}


if(ans>=1)cout<<"Yes"<<endl;
else cout<<"No"<<endl;

return 0;
}