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 MAX_N = 1e5 + 100;
long long n, m;
long long s[MAX_N], c[MAX_N], d[MAX_N];
long long t, l, r;
map<pair<long long, long long>, long long> segs;
vector<pair<pair<long long, long long>, pair<long long, long long> > > queries;
long long incs[MAX_N], maxs[MAX_N];
long long res[MAX_N];
void insert(long long a[], long long x, long long t) {
  for (; x < MAX_N; x += x & (-x)) {
    a[x] += t;
  }
}
long long query(long long a[], long long x) {
  long long ret = 0;
  for (; x; x -= x & (-x)) {
    ret += a[x];
  }
  return ret;
}
long long query(long long a[], long long l, long long r) {
  return query(a, r) - query(a, l - 1);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i] >> c[i] >> d[i];
    segs[make_pair(i, i)] = -1;
    if (d[i] > 0) {
      insert(incs, i, d[i]);
      queries.push_back(
          make_pair(make_pair(c[i] / d[i] + 1, -1), make_pair(i, i)));
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> t >> l >> r;
    while (true) {
      map<pair<long long, long long>, long long>::iterator it =
          segs.upper_bound(make_pair(r, MAX_N));
      if (it == segs.begin()) break;
      it--;
      if (it->first.second < l) {
        break;
      } else if (it->first.second > r) {
        segs[make_pair(it->first.first, r)] = it->second;
        segs[make_pair(r + 1, it->first.second)] = it->second;
      } else if (it->first.first < l) {
        segs[make_pair(it->first.first, l - 1)] = it->second;
        segs[make_pair(l, it->first.second)] = it->second;
      } else if (it->second == -1) {
        int id = it->first.first;
        res[i] += min(c[id], s[id] + d[id] * t);
      } else {
        queries.push_back(make_pair(make_pair(t - it->second, i), it->first));
      }
      segs.erase(it);
    }
    segs[make_pair(l, r)] = t;
  }
  sort(queries.begin(), queries.end());
  for (int i = 0; i < queries.size(); i++) {
    int type = queries[i].first.second;
    t = queries[i].first.first;
    l = queries[i].second.first;
    r = queries[i].second.second;
    if (type == -1) {
      insert(incs, l, -d[l]);
      insert(maxs, l, c[l]);
    } else {
      res[type] += t * query(incs, l, r) + query(maxs, l, r);
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << res[i] << endl;
  }
  return 0;
}