#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    //x=Hまたはy=Wにするまでに必要な操作の回数
    //x=Hにするための操作回数とy=Wにするための操作回数の最小値

    ll N, Q, H, W, P, ok, ng, mid, ans;
    string S;
    cin >> N >> Q >> S;
    vector<ll> D(N+1), R(N+1);
    for (int i=1; i<=N; i++){
        if (S[i-1] == 'D') D[i]++;
        else R[i]++;
        D[i] += D[i-1];
        R[i] += R[i-1];
    }

    //01234
    //DDRRR

    auto f=[&](ll p, ll n, vector<ll> &v){
        n += p;
        ll res=0, q, r;
        //カウンターpから始まってn回押すまでの間に(D or R)がいくつ存在するか？
        q = n / N;
        r = n % N;
        res += v[N] * q;
        res += v[r];
        res -= v[p];
        return res;
    };

    //f(p, X, D) >= Hを満たす最小のX

    for (int i=0; i<Q; i++){
        ans = 1e18;
        cin >> H >> W >> P;
        //Hについて
        if (D[N] != 0){
            ok = N*H; ng = 0;
            while(ok-ng>1){
                mid = (ok+ng)/2;
                if (f(P, mid, D) >= H) ok = mid;
                else ng = mid;
            }
            ans = min(ans, ok);
        }
        //Wについて
        if (R[N] != 0){
            ok = N*W; ng = 0;
            while(ok-ng>1){
                mid = (ok+ng)/2;
                if (f(P, mid, R) >= W) ok = mid;
                else ng = mid;
            }
            ans = min(ans, ok);
        }
        cout << (ans+P)%N << endl;
    }

    return 0;
}