#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, s, n) for (int i = (s); i < (int)(n); i++)
#define rep2(i, n, s) for (int i = (n - 1); i >= (s); i--)
#define all(a)  (a).begin(),(a).end()
#define all_c(a, b) (a).begin(), (a).end(), back_inserter((b))
vector<int> dy = {-1, 0, 1, 0, -1, -1, 1, 1};
vector<int> dx = {0, 1, 0, -1, -1, 1, 1, -1};
const ll INF = numeric_limits<long long>::max();
const ll MOD = 1'000'000'007;
ll unused = INF % MOD;

ll gcd(ll A, ll B) {
    if (B == 0) return A;
    return gcd(B, A % B);
}

ll lcm(ll A, ll B) {
    ll g = gcd(A, B);
    return A / g * B;
}

// 二分検索の雛型
bool binary_search(int N, int A[], int K) {
    int left = 0, right = N - 1;
    while (left <= right) {
        int mid = (left + right) / 2;
        if (A[mid] == K) return true;
        if (A[mid] < K) left = mid + 1;
        else right = mid - 1;
    }
    return false;
}

/*
lower_bound -> if (A[mid] < K)
upper_bound -> if (A[mid] <= K)
*/
int binary_search2(int N, ll A[], ll K) {
    int left = 0, right = N;
    while (left < right) {
        int mid = (left + right) / 2;
        if (A[mid] < K) left = mid + 1;
        else right = mid;
    }
    return right;
}

/* メモ帳
・小数点以下表示 / cout << fixed << setprecision(12) << 
・ルートの計算   / sqrt(A)

・順列全列挙
    do {
        
    } while (next_permutation(V.begin() V.end()));

・bit検索
for (int i = 0; i < (1 << N); i++) {
    rep(j, 0, N) {
        int wari = (1 << j);
        if ((i / wari) % 2 == 1) {
            
        }
    }
}
*/

int N, M;
string S;
vector<vector<int>> V(100009);
ll ans = 0;

int main() {
    cin >> N >> M >> S;
    rep(i, 0, M) {
        int u, v; cin >> u >> v;
        V[u].push_back(v);
        V[v].push_back(u);
    }
    
    queue<int> Q;
    rep(i, 0, N) if (S[i] == 'P') Q.push(i + 1);
    while (Q.size() > 0) {
        int now = Q.front(); Q.pop();
        for (auto next: V[now]) {
            if (S[now - 1] == 'P' && S[next - 1] == 'D') Q.push(next);
            if (S[now - 1] == 'D' && S[next - 1] == 'C') Q.push(next);
            if (S[now - 1] == 'C' && S[next - 1] == 'A') ans++;
        }
    }
    
    cout << ans % 1000000007 << endl;
}