1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
#include <bits/stdc++.h>
using namespace std;
#define fwd(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) fwd(i, 0, n)
#define all(X) X.begin(), X.end()
#define sz(X) int(size(X))
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
using pii = pair<int, int>; using vi = vector<int>;
using ll = long long; using ld = long double;

#ifdef LOC
auto SS = signal(6, [](int) { *(int *)0 = 0; });
#define DTP(x, y) auto operator<<(auto &o, auto a) -> decltype(y, o) { o << "("; x; return o << ")"; }
auto operator<<(auto &o, auto a) -> decltype(all(a), o);
DTP(o << a.fi << ", " << a.se, a.se);
DTP(for (auto i : a) o << i << ", ", all(a));
#define deb(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", [](auto... arg_) { (( cerr << arg_ << ", " ), ...) << '\n'; }(x)
#else
#define deb(...) 0
#endif

const int P = 1e9 + 7, phiP = P - 1;
struct mint {
    int x = 0;
    mint operator+(mint o) const {return {x + o.x >= P ? x + o.x - P : x + o.x}; }
    mint operator-(mint o) const {return {x < o.x ? x - o.x + P : x - o.x}; }
    mint operator+() const {return *this; }
    mint operator-() const {return {x ? P - x : 0}; }
    mint operator*(mint o) const {return {int(ll(x) * o.x % P)}; }
    mint operator/(mint o) const {return *this * o.inv(); }
    mint & operator+=(mint o) {return *this = *this + o; }
    mint & operator-=(mint o) {return *this = *this - o; }
    mint & operator*=(mint o) {return *this = *this * o; }
    mint & operator/=(mint o) {return *this = *this / o; }
    mint pow(ll e) const {
        mint ret{1}, b(*this);
        for (; e; e >>= 1) {
            if (e & 1)
                ret *= b;
            b *= b;
        }
        return ret;
    }
    mint inv() const {return pow(phiP - 1); }
};
auto &operator<<(auto &os, mint m) {
    return os << m.x - (2 * m.x >= P && &os == &cerr ? P : 0);
}

using func = pair<mint, mint>; // ax + b

func operator*(const func &s, const func &t) {
    return {
        s.fi * t.fi,
        s.se + s.fi * t.se
    };
}

func inv(const func &s) {
    mint a = s.fi.inv();
    mint b = -a * s.se;
    func f{a, b};

    {
        func g = f * s;
        assert(g.fi.x == 1);
        assert(g.se.x == 0);
    }

    return f;
}

void solve() {
    int n, q;
    cin >> n >> q;

    vi a(n), b(n); // add, mul

    vector<ll> prefA(n);
    vector<func> prefB(n);
    vi nxt2(n);

    rep(i, n) {
        cin >> a[i] >> b[i];
        prefA[i] = a[i];

        prefB[i].fi.x = b[i] == 1 ? 1 : b[i];
        prefB[i].se.x = b[i] == 1 ? a[i] : 0;

        if (i) {
            prefA[i] += prefA[i-1];
            prefB[i] = prefB[i] * prefB[i-1];
        }
    }

    for (int i = n-1; i >= 0; --i) {
        if (i == n-1)
            nxt2[i] = n;
        else
            nxt2[i] = nxt2[i+1];

        if (b[i] != 1)
            nxt2[i] = i;
    }

    auto sum = [&](int l, int r) -> ll {
        return prefA[r] - (l ? prefA[l-1] : 0);
    };
    auto prod = [&](int l, int r) -> func {
        return prefB[r] * (l ? inv(prefB[l-1]) : func(mint{1}, mint{0}));
    };

    // deb(prefA);
    // deb(prefB);

    rep(_, q) {
        ll x;
        int l, r;
        cin >> x >> l >> r;
        --r;
        while (l <= r) {
            // deb(x, l, r);
            if (x >= P) {
                // deb("skip", l, r);
                auto [A, B] = prod(l, r);
                x = x % P * A.x + B.x;
                break;
            }
            if (b[l] > 1) {
                x = max(x + a[l], x * b[l]);
                ++l;
                continue;
            }
            int R = min(r, nxt2[l] - 1);
            x += sum(l, R);
            l = R + 1;
        }

        x %= P;
        cout << x << '\n';
    }
}

int32_t main() {
    cin.tie(0)->sync_with_stdio(0);
    cout << fixed << setprecision(10);

    int z = 1;
    // cin >> z;
    rep(_, z) solve();

    cout << flush;
    _Exit(0);
}