Problem: Lord Tirek is a centaur and the main antagonist in the season four finale episodes in the series "My Little Pony: Friendship Is Magic". In "Twilight's Kingdom" (Part 1), Tirek escapes from Tartarus and drains magic from ponies to grow stronger.

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The core skill of Tirek is called Absorb Mana. It takes all mana from a magic creature and gives them to the caster.

Now to simplify the problem, assume you have n ponies (numbered from 1 to n). Each pony has three attributes:

  * si : amount of mana that the pony has at time 0; 
  * mi : maximum mana that the pony can have; 
  * ri : mana regeneration per unit time. 



Lord Tirek will do m instructions, each of them can be described with three integers: ti, li, ri. The instruction means that at time ti, Tirek will use Absorb Mana on ponies with numbers from li to ri (both borders inclusive). We'll give you all the m instructions in order, count how much mana Tirek absorbs for each instruction.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of ponies. Each of the next n lines contains three integers si, mi, ri (0 ≤ si ≤ mi ≤ 105; 0 ≤ ri ≤ 105), describing a pony. 

The next line contains an integer m (1 ≤ m ≤ 105) — the number of instructions. Each of the next m lines contains three integers ti, li, ri (0 ≤ ti ≤ 109; 1 ≤ li ≤ ri ≤ n), describing an instruction of Lord Tirek. The instructions are given in strictly increasing order of ti (all ti are distinct).

Output

For each instruction, output a single line which contains a single integer, the total mana absorbed in this instruction.

Examples

Input

5
0 10 1
0 12 1
0 20 1
0 12 1
0 10 1
2
5 1 5
19 1 5


Output

25
58

Note

Every pony starts with zero mana. For the first instruction, each pony has 5 mana, so you get 25 mana in total and each pony has 0 mana after the first instruction.

For the second instruction, pony 3 has 14 mana and other ponies have mana equal to their mi.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct pony {
  int s, mx, r, last;
};
struct info {
  int last;
  vector<tuple<int, int, int>> v;
  map<int, int> sm, sr;
  int allr;
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  const int L = 400;
  int n;
  cin >> n;
  vector<info> b(n / L + 5);
  for (auto &i : b) i.last = -1, i.allr = 0;
  vector<pony> a(n);
  for (auto &i : a) {
    cin >> i.s >> i.mx >> i.r, i.last = -1;
    if (i.r == 0) continue;
    int p = (&i - &a[0]) / L;
    int t = (i.mx + i.r - 1) / i.r;
    b[p].sm[t] += i.mx;
    b[p].sr[t] += i.r;
    b[p].allr += i.r;
  }
  for (auto &i : b) {
    if (!i.sm.count(0)) i.v.push_back({0, 0, i.allr});
    int sm = 0, sr = 0;
    for (auto x : i.sm) {
      sm += x.second;
      sr += i.sr[x.first];
      i.v.push_back({x.first, sm, i.allr - sr});
    }
  }
  auto push = [&](int k) {
    for (int i = k * L; i < (k + 1) * L; ++i)
      a[i].last = max(a[i].last, b[k].last);
    b[k].last = -1;
  };
  auto go = [&](int l, int r, int t) {
    long long ret = 0;
    for (int i = l; i <= r; ++i) {
      if (b[i / L].last != -1) push(i / L);
      if (a[i].last == -1)
        ret += min((long long)a[i].mx, a[i].s + (long long)t * a[i].r);
      else
        ret += min((long long)a[i].mx, (long long)(t - a[i].last) * a[i].r);
      a[i].last = t;
    }
    return ret;
  };
  int q;
  cin >> q;
  for (int t, l, r; q--;) {
    cin >> t >> l >> r;
    --l, --r;
    int bl = l / L;
    int br = r / L;
    if (br - bl < 2) {
      cout << go(l, r, t) << "\n";
      continue;
    }
    long long ans = go(l, bl * L + L - 1, t) + go(br * L, r, t);
    for (int i = bl + 1; i < br; ++i) {
      if (b[i].last == -1) {
        ans += go(i * L, i * L + L - 1, t);
      } else {
        auto pos = lower_bound(b[i].v.begin(), b[i].v.end(),
                               make_tuple(t - b[i].last, 1 << 30, 1 << 30));
        --pos;
        int _t, g, sr;
        tie(_t, g, sr) = *pos;
        ans += g + (long long)(t - b[i].last) * sr;
      }
      b[i].last = t;
    }
    cout << ans << "\n";
  }
  return 0;
}