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 dat {
  int lim;
  long long d, mx;
  dat() {}
  dat(int lim, long long d, long long mx) : lim(lim), d(d), mx(mx) {}
  inline bool operator<(const dat& a) const { return lim < a.lim; }
};
vector<int> s;
struct seg {
  int sz;
  vector<int> tm;
  vector<vector<dat> > d;
  seg() {}
  seg(int mx) {
    sz = 1;
    while (sz <= mx) sz <<= 1;
    tm.resize(sz << 1);
    d.resize(sz << 1);
  }
  void init() {
    for (int i = (sz << 1) - 1; i >= 1; --i) {
      if (i != 1) {
        int p = i >> 1;
        d[p].insert(d[p].end(), d[i].begin(), d[i].end());
      }
      d[i].push_back(dat(0, 0, 0));
      sort(d[i].begin(), d[i].end());
      for (int j = 0; j < (int)d[i].size() - 1; ++j) {
        d[i][j + 1].d += d[i][j].d;
        d[i][j + 1].mx += d[i][j].mx;
      }
    }
  }
  long long sum(int a, int b, int t, int i = 1, int l = 0, int r = -1) {
    if (r == -1) r = sz;
    long long res = 0;
    if (a <= l && r <= b && tm[i]) {
      t -= tm[i];
      int p = lower_bound(d[i].begin(), d[i].end(), dat(t, -1, -1)) -
              d[i].begin() - 1;
      res += (d[i].back().d - d[i][p].d) * t;
      res += d[i][p].mx;
      tm[i] += t;
      return res;
    }
    if (r - l == 1) {
      res = min(d[i].back().d * t + s[i - sz], d[i].back().mx);
      tm[i] = t;
      return res;
    }
    int c = l + r >> 1;
    tm[i << 1] = max(tm[i << 1], tm[i]);
    tm[i << 1 | 1] = max(tm[i << 1 | 1], tm[i]);
    if (a < c) res += sum(a, b, t, i << 1, l, c);
    if (c < b) res += sum(a, b, t, i << 1 | 1, c, r);
    if (a <= l && r <= b)
      tm[i] = t;
    else
      tm[i] = 0;
    return res;
  }
};
int main() {
  int n, q;
  scanf("%d", &n);
  seg d(n);
  s.resize(n);
  for (int i = 0; i < n; ++i) {
    int m, g;
    scanf("%d%d%d", &s[i], &m, &g);
    d.d[d.sz + i].push_back(dat(g ? m / g : 1 << 30, g, m));
  }
  d.init();
  scanf("%d", &q);
  for (int qi = 0; qi < q; ++qi) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    --l;
    printf("%I64d\n", d.sum(l, r, t));
  }
  return 0;
}