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 = 1 << 30;
int n, m, str[100100], lim[100100], up[100100], D[100100], col[100100 * 4];
long long ans;
vector<int> F[100100 * 4];
vector<long long> sum2[100100 * 4], sum[100100 * 4];
bool cmp(int x, int y) { return D[x] > D[y]; }
void build(int node, int l, int r) {
  col[node] = -2;
  if (l == r) {
    F[node].push_back(l);
    sum[node].push_back(up[l]);
    sum2[node].push_back(lim[l]);
    return;
  }
  int mid = (l + r) >> 1, son = node << 1;
  build(son, l, mid);
  build(son | 1, mid + 1, r);
  F[node].resize(r - l + 1);
  sum[node].resize(r - l + 1);
  sum2[node].resize(r - l + 1);
  merge(F[son].begin(), F[son].end(), F[son | 1].begin(), F[son | 1].end(),
        F[node].begin(), cmp);
  sum[node][0] = up[F[node][0]], sum2[node][0] = lim[F[node][0]];
  for (int i = 1, size = F[node].size(); i < size; i++) {
    sum[node][i] = sum[node][i - 1] + up[F[node][i]];
    sum2[node][i] = sum2[node][i - 1] + lim[F[node][i]];
  }
}
void update(int node) {
  if (col[node] > 0) col[node << 1] = col[(node << 1) | 1] = col[node];
}
void getans(int node, int l, int r, int ll, int rr, int t) {
  if (l > rr || ll > r) return;
  if (ll <= l && rr >= r && (col[node] > 0 || col[node] == -2)) {
    if (col[node] > 0) {
      int dif = t - col[node];
      int l = -1, r = F[node].size() - 1;
      while (l < r) {
        int mid = (l + r + 1) >> 1;
        if (D[F[node][mid]] >= dif)
          l = mid;
        else
          r = mid - 1;
      }
      if (l >= 0) ans += sum[node][l] * dif - sum2[node][l];
      ans += sum2[node][F[node].size() - 1];
    } else {
      for (int i = 0, size = F[node].size(); i < size; i++)
        ans += min((long long)lim[F[node][i]],
                   (long long)t * up[F[node][i]] + str[F[node][i]]);
    }
    col[node] = t;
    return;
  }
  update(node);
  int mid = (l + r) >> 1, son = node << 1;
  getans(son, l, mid, ll, rr, t);
  getans(son | 1, mid + 1, r, ll, rr, t);
  if (col[son] == col[son | 1])
    col[node] = col[son];
  else
    col[node] = -1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%d %d %d", &str[i], &lim[i], &up[i]),
        D[i] = up[i] == 0 ? Maxn : lim[i] / up[i];
  build(1, 1, n);
  int t, l, r;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &t, &l, &r);
    ans = 0;
    getans(1, 1, n, l, r, t);
    printf("%lld\n", ans);
  }
}