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 = 1e5 + 200;
int N;
long long S[MAXN];
long long C[MAXN];
long long D[MAXN];
const int MAXM = 1e5 + 200;
int M;
long long T[MAXM];
int L[MAXM];
int R[MAXM];
const int MAXQ = 1e6;
int Q;
pair<pair<int, int>, pair<int, int> > queries[MAXQ];
long long incBit[MAXN];
long long maxBit[MAXN];
long long res[MAXM];
void insert(long long *bit, int a, long long v) {
  a++;
  for (int i = a; i <= N; i += i & (-i)) {
    bit[i] += v;
  }
}
long long query(long long *bit, int a) {
  a++;
  long long v = 0;
  for (int i = a; i > 0; i -= i & (-i)) {
    v += bit[i];
  }
  return v;
}
long long query(long long *bit, int l, int r) {
  return query(bit, r) - query(bit, l - 1);
}
map<pair<int, int>, int> intervals;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++) cin >> S[i] >> C[i] >> D[i];
  cin >> M;
  for (int i = 0; i < M; i++) {
    cin >> T[i] >> L[i] >> R[i];
    L[i]--, R[i]--;
  }
  for (int i = 0; i < N; i++) {
    intervals[pair<int, int>(i, i)] = -1;
  }
  for (int i = 0; i < M; i++) {
    long long t = T[i];
    int l = L[i], r = R[i];
    while (true) {
      auto it = intervals.upper_bound(pair<int, int>(r, N));
      if (it == intervals.begin()) break;
      it--;
      auto v = *it;
      if (v.first.second < l) break;
      if (v.first.second > r) {
        intervals[pair<int, int>(v.first.first, r)] = v.second;
        intervals[pair<int, int>(r + 1, v.first.second)] = v.second;
      } else if (v.first.first < l) {
        intervals[pair<int, int>(v.first.first, l - 1)] = v.second;
        intervals[pair<int, int>(l, v.first.second)] = v.second;
      } else if (v.second == -1) {
        res[i] +=
            min(C[v.first.first], S[v.first.first] + t * D[v.first.first]);
      } else {
        queries[Q++] = make_pair(pair<int, int>(t - v.second, i), v.first);
      }
      intervals.erase(it);
    }
    intervals[pair<int, int>(l, r)] = t;
  }
  memset(incBit, 0, sizeof(incBit));
  memset(maxBit, 0, sizeof(maxBit));
  for (int i = 0; i < N; i++) {
    insert(incBit, i, D[i]);
    if (D[i] > 0)
      queries[Q++] =
          make_pair(pair<int, int>(C[i] / D[i] + 1, -1), pair<int, int>(i, i));
  }
  sort(queries, queries + Q);
  for (int i = 0; i < Q; i++) {
    long long time = queries[i].first.first;
    int type = queries[i].first.second;
    int l = queries[i].second.first;
    int r = queries[i].second.second;
    if (type == -1) {
      insert(incBit, l, -D[l]);
      insert(maxBit, l, C[l]);
    } else {
      res[type] += time * query(incBit, l, r) + query(maxBit, l, r);
    }
  }
  for (int i = 0; i < M; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}