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.

<image>

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;
int dat[300][100001];
long long flag[300];
long long x[100001];
long long now[100000];
long long mx[100000];
long long ver[100000];
long long las[100000];
void init(vector<pair<long long, long long> > vec, int tt) {
  fill(x, x + 100001, 0);
  for (int i = 0; i < vec.size(); i++) {
    if (vec[i].second == 0) continue;
    long long t = vec[i].first / vec[i].second,
              s = vec[i].first - t * vec[i].second;
    x[0] += vec[i].second;
    x[t] -= vec[i].second;
    x[t] += s;
    x[t + 1] -= s;
  }
  for (int i = 1; i <= 100000; i++) x[i] += x[i - 1];
  for (int i = 1; i <= 100000; i++) dat[tt][i] = dat[tt][i - 1] + x[i - 1];
}
void resolve(int b) {
  if (flag[b] == -1) return;
  for (int i = b * 400; i < (b + 1) * 400; i++) {
    now[i] = 0;
    las[i] = flag[b];
  }
  flag[b] = -1;
}
int main() {
  int num;
  scanf("%d", &num);
  for (int i = 0; i < num; i++)
    scanf("%I64d%I64d%I64d", &now[i], &mx[i], &ver[i]);
  for (int i = 0; i < num; i += 400) {
    vector<pair<long long, long long> > z;
    for (int j = i; j < i + 400; j++) z.push_back(make_pair(mx[j], ver[j]));
    init(z, i / 400);
  }
  fill(flag, flag + 300, -1);
  fill(las, las + 100000, 0);
  int query;
  scanf("%d", &query);
  for (int p = 0; p < query; p++) {
    long long zt;
    int zb, zc;
    scanf("%I64d%d%d", &zt, &zb, &zc);
    zb--;
    zc--;
    int a = zb / 400, b = zc / 400;
    resolve(a);
    resolve(b);
    long long ret = 0;
    if (a == b) {
      for (int i = zb; i <= zc; i++) {
        ret += min(mx[i], now[i] + ver[i] * (zt - las[i]));
        now[i] = 0;
        las[i] = zt;
      }
    } else {
      for (int i = zb; i < (a + 1) * 400; i++) {
        ret += min(mx[i], now[i] + ver[i] * (zt - las[i]));
        now[i] = 0;
        las[i] = zt;
      }
      for (int i = a + 1; i <= b - 1; i++) {
        if (flag[i] == -1) {
          for (int j = i * 400; j < (i + 1) * 400; j++) {
            ret += min(mx[j], now[j] + ver[j] * (zt - las[j]));
            now[j] = 0;
          }
        } else {
          ret += dat[i][min(100000LL, zt - flag[i])];
        }
        flag[i] = zt;
      }
      for (int i = b * 400; i <= zc; i++) {
        ret += min(mx[i], now[i] + ver[i] * (zt - las[i]));
        now[i] = 0;
        las[i] = zt;
      }
    }
    printf("%I64d\n", ret);
  }
}