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;
const int N = 100005;
int n, q;
int start[N], max_mana[N], regen[N], need[N];
int minT[N << 2], maxT[N << 2], remS[N << 2], lz[N << 2];
vector<pair<int, int> > vneed[N << 2];
vector<long long> vm[N << 2], vr[N << 2];
void merge(int v) {
  minT[v] = min(minT[v << 1], minT[v << 1 | 1]);
  maxT[v] = max(maxT[v << 1], maxT[v << 1 | 1]);
  remS[v] = remS[v << 1] + remS[v << 1 | 1];
}
void build(int v, int l, int r) {
  vneed[v].push_back(make_pair(-1, 0));
  vm[v].push_back(0);
  vr[v].push_back(0);
  for (int i = l; i <= r; ++i)
    if (regen[i] != 0 && max_mana[i] != 0) {
      vneed[v].push_back(make_pair(need[i], i));
    }
  sort(vneed[v].begin(), vneed[v].end());
  for (int i = 1; i < vneed[v].size(); ++i) {
    vm[v].push_back(vm[v].back() + max_mana[vneed[v][i].second]);
    vr[v].push_back(vr[v].back() + regen[vneed[v][i].second]);
  }
  lz[v] = -1;
  if (l == r) {
    remS[v] = (start[l] > 0);
    return;
  }
  build(v << 1, l, ((l + r) >> 1));
  build(v << 1 | 1, ((l + r) >> 1) + 1, r);
  merge(v);
}
void push(int v, int l, int r) {
  if (lz[v] == -1) return;
  if (l < r) lz[v << 1] = lz[v], lz[v << 1 | 1] = lz[v];
  minT[v] = maxT[v] = lz[v];
  lz[v] = -1;
}
long long query(int v, int l, int r, int L, int R, int t) {
  push(v, l, r);
  if (R < l || L > r) return 0;
  if (L <= l && r <= R) {
    if (remS[v]) {
      if (l == r) {
        remS[v] = 0;
        minT[v] = maxT[v] = t;
        return min(1LL * regen[l] * t + start[l], (long long)max_mana[l]);
      }
    } else {
      if (minT[v] == maxT[v]) {
        int last = minT[v];
        int pos = upper_bound(vneed[v].begin(), vneed[v].end(),
                              make_pair(t - last, N)) -
                  vneed[v].begin();
        long long ret =
            1LL * (t - last) * (vr[v][vneed[v].size() - 1] - vr[v][pos - 1]) +
            vm[v][pos - 1];
        lz[v] = t;
        push(v, l, r);
        return ret;
      }
    }
  }
  long long lef = query(v << 1, l, ((l + r) >> 1), L, R, t);
  long long rig = query(v << 1 | 1, ((l + r) >> 1) + 1, r, L, R, t);
  merge(v);
  return lef + rig;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> start[i] >> max_mana[i] >> regen[i];
    if (regen[i]) need[i] = (max_mana[i] + regen[i] - 1) / regen[i];
  }
  build(1, 1, n);
  cin >> q;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    printf("%lld\n", query(1, 1, n, l, r, t));
  }
}