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 maxn = 262155;
int n, m;
int ps[maxn], pm[maxn], pr[maxn];
int res[maxn];
vector<pair<int, int> > vec[maxn];
vector<long long> pre[maxn], suf[maxn];
void pushUp(int rt) {
  if (res[rt << 1] == res[rt << 1 | 1])
    res[rt] = res[rt << 1];
  else
    res[rt] = -1;
}
void build(int l, int r, int rt) {
  vec[rt] = vector<pair<int, int> >(r - l + 1);
  pre[rt] = suf[rt] = vector<long long>(r - l + 1);
  for (int i = l; i <= r; i++) {
    if (pr[i])
      vec[rt][i - l] = (pair<int, int>((pm[i] + pr[i] - 1) / pr[i], i));
    else
      vec[rt][i - l] = (pair<int, int>(0x3f3f3f3f, i));
  }
  sort(vec[rt].begin(), vec[rt].end());
  long long sp = 0, ss = 0;
  for (int i = 0; i < vec[rt].size(); i++) {
    sp += pm[vec[rt][i].second];
    pre[rt][i] = sp;
  }
  for (int i = int(vec[rt].size()) - 1; i >= 0; i--) {
    ss += pr[vec[rt][i].second];
    suf[rt][i] = ss;
  }
  if (l == r) return;
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void pushDown(int rt) {
  if (res[rt] != -1) {
    res[rt << 1] = res[rt << 1 | 1] = res[rt];
  }
}
long long _solve(int t, int l, int r, int rt) {
  if (res[rt] != -1) {
    if (res[rt] == 0) {
      long long ret = 0;
      for (int i = l; i <= r; i++) {
        ret += min(ps[i] + 1LL * pr[i] * t, 0ll + pm[i]);
      }
      res[rt] = t;
      return ret;
    } else {
      long long ret = 0;
      int ct = t - res[rt];
      int p = int(upper_bound(vec[rt].begin(), vec[rt].end(),
                              pair<int, int>(ct, 0x3f3f3f3f)) -
                  vec[rt].begin());
      if (p > 0) ret += pre[rt][p - 1];
      if (p < vec[rt].size()) ret += 1LL * ct * suf[rt][p];
      res[rt] = t;
      return ret;
    }
  }
  pushDown(rt);
  long long ret = 0;
  int m = (l + r) >> 1;
  ret += _solve(t, l, m, rt << 1);
  ret += _solve(t, m + 1, r, rt << 1 | 1);
  pushUp(rt);
  return ret;
}
long long solve(int L, int R, int t, int l, int r, int rt) {
  if (L <= l && r <= R) {
    return _solve(t, l, r, rt);
  }
  pushDown(rt);
  long long ret = 0;
  int m = (l + r) >> 1;
  if (L <= m) ret += solve(L, R, t, l, m, rt << 1);
  if (R > m) ret += solve(L, R, t, m + 1, r, rt << 1 | 1);
  pushUp(rt);
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &ps[i], &pm[i], &pr[i]);
  scanf("%d", &m);
  build(1, n, 1);
  for (int i = 1; i <= m; i++) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    long long ans = solve(l, r, t, 1, n, 1);
    printf("%lld\n", ans);
  }
  return 0;
}