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#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define st first
#define nd second
typedef long long ll;
typedef long double ld;

const ll I = 1'000'000'000'000'000'000LL;
const int II = 2'000'000'000;
const ll M = 1'000'000'007LL;

pair<int, int> operator*(const pair<int, int> &a, const pair<int, int> &b)
{
    return pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M};
}

pair<int, int> &operator*=(pair<int, int> &a, const pair<int, int> &b)
{
    return (a = pair{((ll)a.st * (ll)b.st) % M, ((ll)a.nd * (ll)b.st + (ll)b.nd) % M});
}

int operator*(const int &a, const pair<int, int> &b)
{
    return ((ll)a * (ll)b.st + (ll)b.nd) % M;
}

int &operator*=(int &a, const pair<int, int> &b)
{
    return (a = ((ll)a * (ll)b.st + (ll)b.nd) % M);
}

const int N = 1<<19;
pair<int, int> drz[2 * N];
pair<int, int> tab[N];
int nxt[N]; ll sum[N];

pair<int, int> Query(int a, int b)
{
    a += N - 1; b += N + 1;
    pair<int, int> res1 = pair{1, 0}, res2 = pair{1, 0};
    while(a / 2 != b / 2)
    {
        if(a % 2 == 0)
            res1 *= drz[a + 1];
        if(b % 2 == 1)
            res2 = (drz[b - 1]) * res2;
        a /= 2; b /= 2;
    }
    return res1 * res2;
}

void Solve()
{
    int n, q, a, b;
    cin >> n >> q;
    for(int i = 1; i <= n; ++i)
    {
        cin >> tab[i].nd >> tab[i].st;
        if(tab[i].st > 1)
            drz[i + N] = pair{tab[i].st, 0};
        else
            drz[i + N] = pair{1, tab[i].nd};
        sum[i] = sum[i - 1] + tab[i].nd;
    }
    for(int v = N - 1; v >= 1; --v)
        drz[v] = drz[v * 2] * drz[v * 2 + 1];
    nxt[n + 1] = n + 1;
    for(int i = n; i >= 1; --i)
    {
        if(tab[i].st >= 2)
            nxt[i] = i;
        else
            nxt[i] = nxt[i + 1];
    }
    for(int i = 1; i <= q; ++i)
    {
        ll ans = 0LL;
        int tmp;
        cin >> tmp >> a >> b;
        ans = tmp;
        ++a;
        while(a <= b && ans < M)
        {
            ans += sum[min(nxt[a] - 1, b)] - sum[a - 1];
            a = nxt[a];
            if(ans < M && a <= b)
            {
                ans = max(ans * (ll)tab[a].st, ans + (ll)tab[a].nd);
                ++a;
            } 
        }
        tmp = ans % M;
        pair<int, int> d = pair{1, 0};
        if(a <= b)
            d = Query(a, b);
        tmp *= d;
        cout << tmp << "\n";
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    //int t; cin >> t;
    //while(t--)
        Solve();

    return 0;
}