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;
struct data {
  int s, m, r, t;
  data(int x = 0) { t = x; }
  void gao() {
    scanf("%d%d%d", &s, &m, &r);
    t = (r ? m / r + (m % r > 0) : 2147483647);
  }
  long long calc(long long u) { return min(s + r * u, (long long)m); }
} a[18][100010];
int Tt[400010];
long long Tm[18][400010], Tr[18][400010];
bool operator<(data a, data b) { return a.t < b.t; }
void build(int l, int r, int x, int d) {
  Tt[x] = -2;
  if (l == r) {
    a[d][l].gao();
    Tm[d][l] = Tm[d][l - 1] + a[d][l].m;
    Tr[d][l] = Tr[d][l - 1] + a[d][l].r;
    return;
  }
  int mid = (l + r) >> 1, i;
  build(l, mid, x << 1, d + 1);
  build(mid + 1, r, x << 1 | 1, d + 1);
  for (i = l; i <= r; i++) a[d][i] = a[d + 1][i];
  sort(a[d] + l, a[d] + r + 1);
  for (i = l; i <= r; i++) {
    Tm[d][i] = Tm[d][i - 1] + a[d][i].m;
    Tr[d][i] = Tr[d][i - 1] + a[d][i].r;
  }
}
void pushdown(int x) {
  if (Tt[x] >= 0) Tt[x << 1] = Tt[x << 1 | 1] = Tt[x];
}
void pushup(int x) {
  Tt[x] = Tt[x << 1];
  if (Tt[x] != Tt[x << 1 | 1]) Tt[x] = -1;
}
long long query(int T, int L, int R, int l, int r, int x, int d) {
  int i;
  long long res = 0;
  if (L <= l && r <= R && Tt[x] != -1) {
    if (Tt[x] == -2) {
      for (i = l; i <= r; i++) res += a[d][i].calc(T);
    } else {
      i = upper_bound(a[d] + l, a[d] + r + 1, data(T - Tt[x])) - a[d] - 1;
      res = (T - Tt[x]) * (Tr[d][r] - Tr[d][i]) + Tm[d][i] - Tm[d][l - 1];
    }
    Tt[x] = T;
    return res;
  }
  pushdown(x);
  i = (l + r) >> 1;
  if (L <= i) res += query(T, L, R, l, i, x << 1, d + 1);
  if (i < R) res += query(T, L, R, i + 1, r, x << 1 | 1, d + 1);
  pushup(x);
  return res;
}
int main() {
  int n, m, t, l, r;
  scanf("%d", &n);
  build(1, n, 1, 0);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &t, &l, &r);
    printf("%I64d\n", query(t, l, r, 1, n, 1, 0));
  }
}