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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 100010, LOG = 20;
long long n, m, k, u, v, x, y, t, l, r, ans;
long long S[MAXN], M[MAXN], R[MAXN], D[MAXN];
int Mn[MAXN << 2], Mx[MAXN << 2], lazy[MAXN << 2];
vector<pair<int, int> > seg[MAXN << 2];
vector<long long> ps1[MAXN << 2], ps2[MAXN << 2];
void Build(int id, int tl, int tr) {
  if (tr - tl == 1) {
    seg[id].push_back({D[tl], tl});
    ps1[id].push_back(0);
    ps1[id].push_back(R[tl]);
    ps2[id].push_back(M[tl]);
    ps2[id].push_back(0);
    return;
  }
  int mid = (tl + tr) >> 1;
  Build(id << 1, tl, mid);
  Build(id << 1 | 1, mid, tr);
  seg[id].resize(tr - tl);
  ps1[id].resize(tr - tl + 1, 0);
  ps2[id].resize(tr - tl + 1, 0);
  merge(seg[id << 1].begin(), seg[id << 1].end(), seg[id << 1 | 1].begin(),
        seg[id << 1 | 1].end(), seg[id].begin());
  ps1[id][0] = ps2[id][tr - tl] = 0;
  for (int i = 0; i < tr - tl; i++)
    ps1[id][i + 1] = ps1[id][i] + M[seg[id][i].second];
  for (int i = tr - tl; i; i--)
    ps2[id][i - 1] = ps2[id][i] + R[seg[id][i - 1].second];
}
void add_lazy(int id, int val) { lazy[id] = Mn[id] = Mx[id] = val; }
void shift(int id) {
  if (!lazy[id]) return;
  add_lazy(id << 1, lazy[id]);
  add_lazy(id << 1 | 1, lazy[id]);
  lazy[id] = 0;
}
void query(int id, int tl, int tr, int l, int r, int t) {
  if (tr <= l || r <= tl) return;
  if (l <= tl && tr <= r && Mn[id] == Mx[id] &&
      (Mn[id] != -1 || tr - tl == 1)) {
    long long time = t - Mn[id];
    int shit = (Mn[id] == -1);
    add_lazy(id, t);
    if (tr - tl == 1) {
      time -= shit;
      ans += min(M[tl], S[tl] + R[tl] * time);
      S[tl] = 0;
      return;
    }
    int pos = upper_bound(seg[id].begin(), seg[id].end(),
                          pair<int, int>(time, inf + 1)) -
              seg[id].begin();
    ans += ps1[id][pos];
    ans += ps2[id][pos] * time;
    return;
  }
  shift(id);
  int mid = (tl + tr) >> 1;
  query(id << 1, tl, mid, l, r, t);
  query(id << 1 | 1, mid, tr, l, r, t);
  Mn[id] = min(Mn[id << 1], Mn[id << 1 | 1]);
  Mx[id] = max(Mx[id << 1], Mx[id << 1 | 1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(Mn, -1, sizeof(Mn));
  memset(Mx, -1, sizeof(Mx));
  cin >> n;
  for (int i = 1; i <= n; i++)
    cin >> S[i] >> M[i] >> R[i], D[i] = (R[i] ? (M[i] + R[i] - 1) / R[i] : inf);
  Build(1, 1, n + 1);
  cin >> m;
  while (m--) {
    cin >> t >> l >> r;
    ans = 0;
    query(1, 1, n + 1, l, r + 1, t);
    cout << ans << '\n';
  }
  return 0;
}