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;
vector<pair<int, pair<long long, long long> > > t[1 << 20];
int cur[1 << 20];
int state[1 << 20];
int lll[1 << 20];
pair<int, int> s[1 << 20];
void build(int p, int u, int node) {
  state[node] = 1;
  cur[node] = 0;
  lll[node] = -1;
  if (p == u)
    t[node].push_back(make_pair(
        s[p].second ? (s[p].first + s[p].second - 1) / s[p].second : (1 << 30),
        make_pair(s[p].first, s[p].second)));
  else {
    int m = (p + u) / 2;
    build(p, m, node << 1);
    build(m + 1, u, node << 1 | 1);
    t[node].resize(t[node << 1].size() + t[node << 1 | 1].size());
    merge(t[node << 1].begin(), t[node << 1].end(), t[node << 1 | 1].begin(),
          t[node << 1 | 1].end(), t[node].begin());
  }
}
void go(int p, int u, int node) {
  int sz = t[node].size();
  for (int i = 1; i < sz; ++i)
    t[node][i].second.first += t[node][i - 1].second.first;
  for (int i = sz - 2; i >= 0; --i)
    t[node][i].second.second += t[node][i + 1].second.second;
  if (p == u) return;
  int m = (p + u) / 2;
  go(p, m, node << 1);
  go(m + 1, u, node << 1 | 1);
}
void lazy(int p, int u, int node) {
  if (lll[node] == -1) return;
  cur[node] = lll[node];
  if (p != u) {
    lll[node << 1] = lll[node << 1 | 1] = lll[node];
    state[node] = 1;
  }
  lll[node] = -1;
}
long long Query(int p, int u, int l, int r, int Time, int node) {
  long long ans = 0;
  lazy(p, u, node);
  if (l <= p && u <= r && state[node]) {
    int where = lower_bound(t[node].begin(), t[node].end(),
                            make_pair(Time - cur[node] + 1, make_pair(-1, -1)),
                            less<pair<int, pair<long long, long long> > >()) -
                t[node].begin();
    int sz = t[node].size();
    if (where < sz)
      ans += 1ll * (Time - cur[node]) * t[node][where].second.second;
    --where;
    if (where >= 0) ans += t[node][where].second.first;
  } else {
    int m = (p + u) / 2;
    if (l <= m) ans += Query(p, m, l, r, Time, node << 1);
    if (m + 1 <= r) ans += Query(m + 1, u, l, r, Time, node << 1 | 1);
  }
  if (l <= p && u <= r) {
    lll[node] = Time;
    lazy(p, u, node);
  } else {
    int m = (p + u) / 2;
    lazy(p, m, node << 1);
    lazy(m + 1, u, node << 1 | 1);
    state[node] = state[node << 1] && state[node << 1 | 1] &&
                  cur[node << 1] == cur[node << 1 | 1];
  }
  return ans;
}
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  static int now[1 << 20];
  int n;
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  set<int> S;
  for (int i = 1; i <= n; ++i) S.insert(i);
  for (int i = 1; i <= n; ++i) cin >> now[i] >> s[i].first >> s[i].second;
  build(1, n, 1);
  go(1, n, 1);
  int m;
  cin >> m;
  int was = 0;
  while (m--) {
    int a, b, c;
    cin >> a >> b >> c;
    long long ans = 0;
    while (1) {
      auto it = S.lower_bound(b);
      if (it == S.end()) break;
      if (*it > c) break;
      ans += min(1ll * s[*it].first, 1ll * a * s[*it].second + now[*it]) -
             min(1ll * s[*it].first, 1ll * a * s[*it].second);
      S.erase(it);
    }
    ans += Query(1, n, b, c, a, 1);
    was = a;
    cout << ans << '\n';
  }
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}