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>
int N, Q;
std::vector<int> V[262144];
std::vector<long long> SM[262144], SR[262144];
long long S[100001], M[100001], R[100001], Time[262144];
bool Same[262144];
inline bool cmp(int x, int y) {
  if (!M[y]) return false;
  if (!M[x]) return true;
  if (!R[x]) return false;
  if (!R[y]) return true;
  return M[x] * R[y] < M[y] * R[x];
}
void build(int p, int l, int r) {
  if (l == r) {
    V[p] = {l};
    SM[p] = {0, M[l]};
    SR[p] = {R[l], 0};
    Same[p] = 1;
    Time[p] = -1;
    return;
  }
  int m = l + r >> 1;
  build(p << 1, l, m);
  build(p << 1 | 1, m + 1, r);
  V[p].clear();
  V[p].reserve(V[p << 1].size() + V[p << 1 | 1].size());
  for (auto i = V[p << 1].begin(), j = V[p << 1 | 1].begin();
       i != V[p << 1].end() || j != V[p << 1 | 1].end();)
    V[p].push_back(
        *(i == V[p << 1].end()
              ? j
              : (j == V[p << 1 | 1].end() ? i : (cmp(*i, *j) ? i : j)))++);
  SM[p].resize(V[p].size() + 1);
  SR[p].resize(V[p].size() + 1);
  size_t S = V[p].size();
  int *v = V[p].data();
  long long *sm = SM[p].data(), *sr = SR[p].data();
  for (size_t i = 1; i <= S; i++) sm[i] = sm[i - 1] + M[v[i - 1]];
  for (size_t i = S - 1; ~i; i--) sr[i] = sr[i + 1] + R[v[i]];
  Same[p] = false;
}
long long G(int p, int l, int r, int L, int R, long long TIME) {
  if (Same[p] && L <= l && r <= R) {
    size_t al = 0, ar = V[p].size();
    while (al < ar) {
      size_t am = al + ar >> 1;
      if (::R[V[p][am]] * (TIME - Time[p]) < M[V[p][am]])
        ar = am;
      else
        al = am + 1;
    }
    return Time[p] == -1 ? std::min(M[l], S[l] + TIME * ::R[l])
                         : SM[p][al] + SR[p][al] * (TIME - Time[p]);
  }
  int m = l + r >> 1;
  if (Same[p]) {
    Same[p << 1] = Same[p << 1 | 1] = true;
    Time[p << 1] = Time[p << 1 | 1] = Time[p];
  }
  if (R <= m) return G(p << 1, l, m, L, R, TIME);
  if (L > m) return G(p << 1 | 1, m + 1, r, L, R, TIME);
  return G(p << 1, l, m, L, R, TIME) + G(p << 1 | 1, m + 1, r, L, R, TIME);
}
void P(int p, int l, int r, int L, int R, long long TIME) {
  if (L <= l && r <= R) {
    Same[p] = true;
    Time[p] = TIME;
    return;
  }
  int m = l + r >> 1;
  if (Same[p]) {
    Same[p << 1] = Same[p << 1 | 1] = 1;
    Time[p << 1] = Time[p << 1 | 1] = Time[p];
  }
  Same[p] = false;
  if (L <= m) P(p << 1, l, m, L, R, TIME);
  if (R > m) P(p << 1 | 1, m + 1, r, L, R, TIME);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld%lld%lld", S + i, M + i, R + i);
  build(1, 1, N);
  for (scanf("%d", &Q); Q--;) {
    long long t;
    int l, r;
    scanf("%lld%d%d", &t, &l, &r);
    printf("%lld\n", G(1, 1, N, l, r, t));
    P(1, 1, N, l, r, t);
  }
  return 0;
}