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#include<bits/stdc++.h>

using namespace std;

constexpr int mod = 1000000007;

constexpr int czer = 0;
constexpr int ziel = 1;
constexpr int nieb = 2;

int color_of(char x) {
    if (x == 'C') return czer;
    if (x == 'Z') return ziel;
    return nieb;
}

long long power(int a, int e) {
    if (e == 0) return 1;
    if (e % 2 == 1) return (a * power(a, e - 1)) % mod;
    long long t = power(a, e / 2);
    return t * t % mod;
}

// 3^{-1} modulo 1000000007
constexpr int inv_3 = 333333336;

int tab[6][3] = {
    {1, 0, 0},  // 1 0 0
    {1, 1, 0},  // 1 1 0
    {0, 1, 0},  // 1 2 1
    {0, 1, 1},  // 2 3 3 
    {0, 0, 1},  // 5 5 6
    {1, 0, 1},  // 11 10 11
};

long long sum_3(int n, int r) {
    int pow_2 = power(2, n);
    if (n % 2 == 0) 
        pow_2 = (pow_2 + mod - 1) % mod;
    else
        pow_2 = (pow_2 + mod - 2) % mod;
    return (((long long)(pow_2) * inv_3 + tab[n % 6][r])) % mod;    
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    
    int n, q; cin >> n >> q;
    string s; cin >> s;

    vector<int> in(n);
    int cnt[3] {0, 0, 0};
    for (int i = 0; i < n; i++) {
        in[i] = color_of(s[i]);
        cnt[in[i]]++;
    }

    for (int query = 0; query <= q; query++) {
        if (query >= 1) {
            int id; char ch;
            cin >> id >> ch;
            int old_col = in[id - 1];
            int new_col = color_of(ch);
            cnt[old_col]--;
            cnt[new_col]++;
            in[id - 1] = new_col;
        }
        if (cnt[nieb] == 0) {
            int ans((cnt[ziel] % 3) != (cnt[czer] % 3));
            for (int d : {0, 1}) {
                bool good = (n >= 3); 
                for (int i = 0; i < n; i++)
                    good &= (((in[i] + i) % 2) == d);
                ans -= int(good);
            }
            cout << ans << "\n";
            continue;
        }
        // initially, assume all are good
        int ans = power(2, cnt[nieb]);
        if (n >= 3 and n % 2 == 1 and cnt[nieb] > 0) {
            for (int d : {0, 1}) {
                bool good = true;
                for (int i = 0; i < n; i++)
                    good &= (in[i] == nieb or ((in[i] + i) % 2) == d);
                if (good) ans = (ans + mod - 1) % mod;
            }
        }
        int dif = (3 + cnt[czer] - (cnt[ziel] % 3)) % 3;
        int r = (3 + 2 * dif - (cnt[nieb] % 3)) % 3;
        cout << (mod + ans - sum_3(cnt[nieb], r)) % mod << "\n";
    }
}