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>
const int N = 100005;
int n, m, max;
struct {
  int s, m, r;
} a[N];
bool start[N];
long long ans[N];
struct {
  long long a[N], b[N];
  void modify(int l, int r, int v) {
    for (int i = l; i <= max; i += i & -i) a[i] += v, b[i] += (long long)v * l;
    for (int i = r + 1; i <= max; i += i & -i)
      a[i] -= v, b[i] -= (long long)v * (r + 1);
  }
  long long query(int r) {
    long long ans = 0;
    r = std::min(r, max);
    for (int i = r; i; i &= i - 1) ans += (r + 1) * a[i] - b[i];
    return ans;
  }
} bit;
std::map<int, int> map;
std::map<int, int>::iterator split(int pos) {
  std::map<int, int>::iterator it = map.lower_bound(pos);
  if (it->first == pos) return it;
  return map.emplace(pos, (--it)->second).first;
}
std::vector<std::tuple<int, int, int>> query[N];
void assign(int l, int r, int time, int id) {
  std::map<int, int>::iterator L = split(l), R = split(r + 1);
  for (std::map<int, int>::iterator i = L; i != R; ++i) {
    int p = i->first, q = std::next(i)->first - 1;
    if (!start[p]) {
      ans[id] += std::min<long long>(a[p].s + (long long)time * a[p].r, a[p].m);
      start[p] = 1;
    } else {
      query[p - 1].emplace_back(time - i->second, -1, id);
      query[q].emplace_back(time - i->second, 1, id);
    }
  }
  map.erase(L, R), map[l] = time;
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n, map[n + 1] = -233;
  for (int i = 1; i <= n; ++i) {
    std::cin >> a[i].s >> a[i].m >> a[i].r;
    if (a[i].r) max = std::max(max, a[i].m / a[i].r + 1);
  }
  for (int i = 1; i <= n; ++i) map[i] = 0;
  std::cin >> m;
  for (int i = 1; i <= m; ++i) {
    int t, l, r;
    std::cin >> t >> l >> r;
    assign(l, r, t, i);
  }
  for (int i = 1; i <= n; ++i) {
    int p = a[i].m / a[i].r;
    if (a[i].r) {
      bit.modify(1, p, a[i].r);
      bit.modify(p + 1, p + 1, a[i].m % a[i].r);
    }
    for (auto q : query[i]) {
      ans[std::get<2>(q)] += std::get<1>(q) * bit.query(std::get<0>(q));
    }
  }
  for (int i = 1; i <= m; ++i) std::cout << ans[i] << '\n';
  return 0;
}