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;
int n, k, q, t, x, y, s;
int last[100005], mx[100005], reg[100005], arr[100005];
long long bas[405][405], son[405][405], ok[405];
vector<pair<double, int> > V[405];
long long calc(int x, int b, int s, int t) {
  int i;
  long long sum = 0;
  if (ok[x] == -1) {
    if (s - b + 1 == k) {
      for (i = b; i <= s; i++) {
        sum +=
            min((long long)mx[i], reg[i] * (long long)(t - arr[i]) + last[i]);
        arr[i] = t;
        last[i] = 0;
      }
      ok[x] = t;
      return sum;
    }
    for (i = b; i <= s; i++) {
      sum += min((long long)(t - arr[i]) * reg[i] + last[i], (long long)mx[i]);
      last[i] = 0;
      arr[i] = t;
    }
    return sum;
  }
  if (s - b + 1 == k) {
    int p =
        upper_bound(V[x].begin(), V[x].end(), make_pair((double)t - ok[x], 0)) -
        V[x].begin();
    sum = (!!p) * bas[x][p - 1] + ((p < V[x].size()) * son[x][p] * (t - ok[x]));
    ok[x] = t;
    return sum;
  }
  for (i = (x - 1) * k + 1; i <= x * k; i++) {
    if (i < b or i > s) {
      arr[i] = ok[x];
    } else {
      sum += min((long long)(t - ok[x]) * reg[i] + last[i], (long long)mx[i]);
      last[i] = 0;
      arr[i] = t;
    }
  }
  ok[x] = -1;
  return sum;
}
int main() {
  cin >> n;
  k = sqrt(n);
  int i, j;
  for (i = 1; i <= n; i++) scanf("%d %d %d", last + i, mx + i, reg + i);
  for (i = 1; i <= n; i += k) {
    int s = i / k + 1;
    for (j = i; j < i + k; j++)
      if (!reg[j])
        V[s].push_back(make_pair(10000000.0, j));
      else
        V[s].push_back(make_pair((double)mx[j] / reg[j], j));
    sort(V[s].begin(), V[s].end());
    for (j = 0; j < k; j++) bas[s][j] = mx[V[s][j].second] + bas[s][j - 1];
    for (j = k - 1; j >= 0; j--)
      son[s][j] = son[s][j + 1] + reg[V[s][j].second];
  }
  memset(ok, -1, sizeof ok);
  cin >> q;
  for (i = 1; i <= q; i++) {
    scanf("%d %d %d", &t, &x, &y);
    long long ans = 0;
    for (j = 1; j <= y; j += k)
      if (x < j + k) ans += calc(j / k + 1, max(j, x), min(j + k - 1, y), t);
    printf("%lld\n", ans);
  }
  return 0;
}