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 maxn = 1e5 + 379, S = 320;
vector<int> blocktime(S, -1), idtime(maxn, 0), mana(maxn), maxi(maxn), re(maxn);
long long sum[S][maxn];
void unlazy(int b) {
  for (int i = b * S; i < (b + 1) * S; i++)
    if (blocktime[b] > -1) idtime[i] = blocktime[b];
  blocktime[b] = -1;
}
void update(int li, int ri, int t, long long &ans) {
  for (int i = li; i <= ri; i++) {
    ans += min((long long)maxi[i], (t - idtime[i]) * 1ll * re[i] + mana[i]);
    mana[i] = 0;
    idtime[i] = t;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> mana[i] >> maxi[i] >> re[i];
    if (!re[i]) continue;
    sum[i / S][1] += re[i];
    sum[i / S][maxi[i] / re[i] + 1] += maxi[i] % re[i] - re[i];
    sum[i / S][maxi[i] / re[i] + 2] -= maxi[i] % re[i];
  }
  for (int i = 0; i < S; i++) {
    for (int j = 1; j < maxn; j++) sum[i][j] += sum[i][j - 1];
    for (int j = 1; j < maxn; j++) sum[i][j] += sum[i][j - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    l--, r--;
    long long ans = 0;
    unlazy(l / S);
    if (l / S == r / S)
      update(l, r, t, ans);
    else {
      unlazy(r / S);
      update(l, (l / S + 1) * S - 1, t, ans);
      update((r / S) * S, r, t, ans);
      for (int i = l / S + 1; i < r / S; i++) {
        if (blocktime[i] == -1)
          update(i * S, (i + 1) * S - 1, t, ans);
        else
          ans += sum[i][min(maxn - 1, t - blocktime[i])];
        blocktime[i] = t;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}