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;
const int N = 100005, M = 1005;
int m[N], R[N], t[N], f[N], s[N], flag[M], n, q, T, x, y, S, len[M], val[M][M];
long long v[N], pre[M][M], suf[M][M];
int cmp(int x, int y) { return t[x] < t[y]; }
void down(int x) {
  if (flag[x] != -1)
    for (int i = (x - 1) * S + 1; i <= n && i <= x * S; i++)
      v[i] = -(long long)flag[x] * R[i];
  flag[x] = -1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d", &s[i], &m[i], &R[i]),
        t[i] = R[i] == 0 ? 1e9 + 1 : m[i] / R[i], f[i] = i, v[i] = s[i];
  S = 1000;
  for (int i = 1; i <= (n - 1) / S + 1; i++) {
    flag[i] = -1;
    int l = (i - 1) * S + 1, r = min(n, i * S);
    len[i] = r - l + 1;
    sort(f + l, f + r + 1, cmp);
    for (int j = 1; j <= len[i]; j++) {
      val[i][j] = t[f[l + j - 1]];
      pre[i][j] = pre[i][j - 1] + m[f[l + j - 1]];
    }
    for (int j = len[i]; j; j--) suf[i][j] = suf[i][j + 1] + R[f[l + j - 1]];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d", &T, &x, &y);
    int l = (x - 1) / S + 1, r = (y - 1) / S + 1;
    long long ans = 0;
    if (l == r) {
      down(l);
      for (int i = x; i <= y; i++)
        ans += min((long long)m[i], v[i] + (long long)R[i] * T),
            v[i] = -(long long)R[i] * T;
    } else {
      down(l);
      down(r);
      for (int i = x; i <= l * S; i++)
        ans += min((long long)m[i], v[i] + (long long)R[i] * T),
            v[i] = -(long long)R[i] * T;
      for (int i = (r - 1) * S + 1; i <= y; i++)
        ans += min((long long)m[i], v[i] + (long long)R[i] * T),
            v[i] = -(long long)R[i] * T;
      for (int i = l + 1; i < r; i++) {
        if (flag[i] != -1) {
          int vv = T - flag[i];
          int L = 0, R = len[i];
          while (L < R) {
            int mid = (L + R + 1) / 2;
            if (val[i][mid] < vv)
              L = mid;
            else
              R = mid - 1;
          }
          ans += pre[i][L] + suf[i][L + 1] * vv;
        } else {
          for (int j = (i - 1) * S + 1; j <= i * S; j++)
            ans += min((long long)m[j], v[j] + (long long)R[j] * T),
                v[j] = -(long long)R[j] * T;
        }
        flag[i] = T;
      }
    }
    printf("%lld\n", ans);
  }
}