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 _ {
  int m, r;
  long long M, R;
  bool operator<(const _ &o) const {
    if (r == 0 && o.r == 0)
      return m < o.m;
    else if (r == 0)
      return false;
    else if (o.r == 0)
      return true;
    return m / r < o.m / o.r;
  }
};
int s[200000], m[200000], r[200000];
int lt[200000];
int blt[200000];
_ x[200000];
int n, Z;
long long min(long long a, long long b) {
  return (((a - b) >> 63) & (a ^ b)) ^ b;
}
void read(int &r) {
  r = 0;
  char ch;
  while ((ch = getchar()) < '0')
    ;
  while (ch >= '0') {
    r = r * 10 + ch - '0';
    ch = getchar();
  }
}
int main() {
  read(n);
  for (int i = (0); i < (n); i++) read(s[i]), read(m[i]), read(r[i]);
  for (int i = (0); i < (n); i++) lt[i] = 0, blt[i] = -1;
  for (int i = (0); i < (n); i++)
    x[i].m = x[i].M = m[i], x[i].r = x[i].R = r[i];
  for (int i = 0; i < n; i += 5375) {
    int S = i, E = min(i + 5375, n);
    sort(x + S, x + E);
    for (int i = (S + 1); i < (E); i++) x[i].m += x[i - 1].m;
    for (int i = E - 2; i >= S; i--) x[i].r += x[i + 1].r;
  }
  read(Z);
  for (int T = (0); T < (Z); T++) {
    int _t, L, R;
    read(_t);
    read(L);
    read(R);
    L--;
    R--;
    long long res = 0, t = _t;
    int bl = L / 5375, br = R / 5375;
    if (bl == br) {
      if (blt[bl] != -1) {
        for (int i = (bl * 5375); i < (bl * 5375 + 5375); i++) lt[i] = blt[bl];
      }
      blt[bl] = -1;
      for (int i = (L); i < (R + 1); i++) {
        res += min(m[i], s[i] + r[i] * (t - lt[i]));
        s[i] = 0;
        lt[i] = t;
      }
    } else {
      if (blt[bl] != -1) {
        for (int i = (bl * 5375); i < (bl * 5375 + 5375); i++) lt[i] = blt[bl];
      }
      if (blt[br] != -1) {
        for (int i = (br * 5375); i < (br * 5375 + 5375); i++) lt[i] = blt[br];
      }
      blt[bl] = -1;
      for (int i = (L); i < (bl * 5375 + 5375); i++) {
        res += min(m[i], s[i] + r[i] * (t - lt[i]));
        s[i] = 0;
        lt[i] = t;
      }
      blt[br] = -1;
      for (int i = (br * 5375); i < (R + 1); i++) {
        res += min(m[i], s[i] + r[i] * (t - lt[i]));
        s[i] = 0;
        lt[i] = t;
      }
      for (int bi = (bl + 1); bi < (br); bi++) {
        if (blt[bi] == -1) {
          for (int i = (bi * 5375); i < (bi * 5375 + 5375); i++) {
            res += min(m[i], s[i] + r[i] * (t - lt[i]));
            s[i] = 0;
            lt[i] = t;
          }
          blt[bi] = t;
        } else {
          int S = bi * 5375, E = min(bi * 5375 + 5375, n);
          int lastcap = S - 1;
          int s = S, e = E - 1;
          while (s <= e) {
            int md = (s + e) / 2;
            if ((t - blt[bi]) * x[md].R > x[md].M) {
              lastcap = md;
              s = md + 1;
            } else
              e = md - 1;
          }
          if (lastcap != S - 1) res += x[lastcap].m;
          if (lastcap + 1 != E) res += x[lastcap + 1].r * (t - blt[bi]);
          blt[bi] = t;
        }
      }
    }
    printf("%lld\n", res);
  }
}