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;
long long s[100500], r[100500], m[100500];
struct st {
  long long lifeTime;
  long long maxMana;
  long long increment;
  st(long long lifeTime, long long maxMana, long long increment)
      : lifeTime(lifeTime), maxMana(maxMana), increment(increment) {}
};
vector<st> vt[4 * 100500];
long long lastTime[4 * 100500];
bool comp(long long t, const st &a) { return t < a.lifeTime; }
bool comp2(const st &a, const st &b) { return a.lifeTime < b.lifeTime; }
void maketree(int node, int left, int right) {
  for (int i = left; i <= right; i++) {
    long long tem = r[i] ? (m[i] + r[i] - 1ll) / r[i] : 1e18 + 1;
    vt[node].push_back(st(tem, m[i], r[i]));
  }
  sort(vt[node].begin(), vt[node].end(), comp2);
  for (int i = 1; i < (int)vt[node].size(); i++) {
    vt[node][i].maxMana += vt[node][i - 1].maxMana;
    vt[node][i].increment += vt[node][i - 1].increment;
  }
  if (left == right) {
    return;
  }
  int l = node << 1;
  int r = l | 1;
  int mid = (left + right) >> 1;
  maketree(l, left, mid);
  maketree(r, mid + 1, right);
}
long long query(int node, int left, int right, int a, int b,
                long long curTime) {
  if (left > b || right < a) return 0ll;
  if (left >= a && right <= b && lastTime[node]) {
    long long ret = vt[node].back().increment * (curTime - lastTime[node]);
    if (vt[node][0].lifeTime <= curTime - lastTime[node]) {
      int idx = upper_bound(vt[node].begin(), vt[node].end(),
                            curTime - lastTime[node], comp) -
                vt[node].begin();
      ret -= vt[node][idx - 1].increment * (curTime - lastTime[node]);
      ret += vt[node][idx - 1].maxMana;
    }
    lastTime[node] = curTime;
    return ret;
  }
  if (left == right) {
    long long ret = min(m[left], s[left] + r[left] * (curTime));
    s[left] = 0ll;
    lastTime[node] = curTime;
    return ret;
  }
  int l = node << 1;
  int r = l | 1;
  int mid = (left + right) >> 1;
  if (lastTime[node]) {
    lastTime[l] = max(lastTime[l], lastTime[node]);
    lastTime[r] = max(lastTime[r], lastTime[node]);
  }
  long long ret = query(l, left, mid, a, b, curTime) +
                  query(r, mid + 1, right, a, b, curTime);
  lastTime[node] = 0ll;
  if (lastTime[l] == lastTime[r]) lastTime[node] = lastTime[l];
  return ret;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld %lld %lld", &s[i], &m[i], &r[i]);
  }
  maketree(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b;
    long long curTime;
    scanf("%lld", &curTime);
    scanf("%d %d", &a, &b);
    printf("%lld\n", query(1, 1, n, a, b, curTime));
    ;
  }
  return 0;
}