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 std::max;
using std::min;
int s[100010], m[100010], r[100010], n, q;
int lson[10000001], rson[10000001], root[100010], T[100010], R[100010];
long long sumM[10000001], sumR[10000001], cnt, ans;
std::set<int> set;
void add(int &pos, int pre, int l, int r, int e, int x, int y) {
  pos = ++cnt, lson[pos] = lson[pre], rson[pos] = rson[pre],
  sumM[pos] = sumM[pre] + x, sumR[pos] = sumR[pre] + y;
  if (l == r) return;
  if ((l + r) >> 1 >= e)
    add(lson[pos], lson[pre], l, (l + r) >> 1, e, x, y);
  else
    add(rson[pos], rson[pre], ((l + r) >> 1) + 1, r, e, x, y);
}
long long queryright(int pos, int pre, int l, int r, int e) {
  if (!pos) return 0;
  if (l >= e) return sumR[pos] - sumR[pre];
  if (r < e) return 0;
  return queryright(lson[pos], lson[pre], l, (l + r) >> 1, e) +
         queryright(rson[pos], rson[pre], ((l + r) >> 1) + 1, r, e);
}
long long queryleft(int pos, int pre, int l, int r, int e) {
  if (!pos) return 0;
  if (r <= e) return sumM[pos] - sumM[pre];
  if (l > e) return 0;
  return queryleft(lson[pos], lson[pre], l, (l + r) >> 1, e) +
         queryleft(rson[pos], rson[pre], ((l + r) >> 1) + 1, r, e);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", s + i, m + i, r + i);
    if (r[i])
      add(root[i], root[i - 1], 0, 100000, m[i] / r[i], m[i], r[i]);
    else
      root[i] = root[i - 1];
    set.insert(i);
    T[i] = 0;
  }
  scanf("%d", &q);
  set.insert(n + 1);
  for (int i = 1, t, l, r; i <= q; i++) {
    scanf("%d%d%d", &t, &l, &r);
    long long ans = 0;
    int a = -1, b = -1, L, t1, t2;
    for (auto x = --set.upper_bound(l); *x <= r && x != set.end();
         set.erase(L)) {
      L = *x;
      t1 = max(l, L);
      t2 = min(r, *(++x) - 1);
      if (!T[L])
        ans += min((long long)::r[t1] * t + s[t1], (long long)m[t1]);
      else
        ans += queryright(root[t2], root[t1 - 1], 0, 100000, t - T[L]) *
                   (t - T[L]) +
               queryleft(root[t2], root[t1 - 1], 0, 100000, t - T[L] - 1);
      if (t2 < *x - 1) T[r + 1] = T[L], b = r + 1;
      if (t1 > L) a = L;
    }
    set.insert(l);
    if (~a) set.insert(a);
    if (~b) set.insert(b);
    T[l] = t;
    printf("%lld\n", ans);
  }
}