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#include <bits/stdc++.h>

#define ll long long
#define pi std::pair<int, int>
#define pll std::pair<ll, ll>
#define vi std::vector<int>
#define vll std::vector<ll>
#define vpi std::vector<pi>
#define vpll std::vector<pll>
#define si std::set<int>

// logika: moge przejsc max 20 razy, by zwiekszyc liczbe zolnierzy tak, by oplacalo sie tylko mnozyc
// z wyjatkiem ze musze jakos obsluzyc zapytania typu (a, 1)

// moge dodac jakis balans a potem robic po tym autostrade, tzn. miec fixed sume, oraz miec tablice
// na bazie ktorej szybko przemnoze?

ll x = 1, y = 0, last_y; // przy kazdym gcd na start x = 1 i y = 0; inaczej sie rozjezdza
ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    ll _gcd = gcd(b, a % b);
    last_y = y;
    y = x - a / b * last_y;
    x = last_y;
    return _gcd;
}

void solve() {
    ll n, q, mod = 1e9 + 7;
    std::cin >> n >> q;

    vll multtab(n+1), sumtab(n+1), jump(n+1), prefsum(n+1);
    multtab[0] = 1;
    vpll v(n+1);
    for(int i = 1; i <= n; i++) {
        std::cin >> v[i].first >> v[i].second;

        sumtab[i] = sumtab[i-1];
        if(v[i].second == 1)
            sumtab[i] += v[i].first;
        else
            sumtab[i] *= v[i].second;
        sumtab[i] %= mod;

        prefsum[i] = prefsum[i-1] + v[i].first;

        multtab[i] = multtab[i-1] * v[i].second;
        multtab[i] %= mod;
    }

    ll last = n+1;
    for(int i = n; i >= 0; i--) {
        jump[i] = last;
        if(v[i].second != 1)
            last = i;
    }
 
    ll ox, l, r; 
    for(int i = 0; i < q; i++) {
        std::cin >> ox >> l >> r;

        ll nl = l + 1, nx = ox;
        while(nl <= r && nx < mod) {// idk chyba tak
            if(v[nl].second == 1) {
                nx += prefsum[std::min(jump[nl]-1, r)] - prefsum[nl-1];
                nl = std::min(jump[nl]-1, r) + 1;
            }
            else {
                nx = std::max(nx + v[nl].first, nx * v[nl].second);
                nl++;
            }
        }
        nx %= mod;
        
        nx -= sumtab[nl-1];
        if(nx < 0)
            nx += mod;

        nx *= multtab[r];
        nx %= mod;

        x = 1, y = 0;
        gcd(multtab[nl-1], mod);
        ll inv = (x % mod + mod) % mod;
        nx *= inv;
        nx %= mod;

        nx += sumtab[r];
        nx %= mod;

        std::cout << nx << '\n';
    }

    
}

int main() {
    std::ios_base::sync_with_stdio(0);
    std::cin.tie(0);

    int t = 1;
    //std::cin >> t;
    while(t--) {
        solve();
    }
}