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 = 100005;
int s[MAXN], lim[MAXN], reg[MAXN], m, n;
vector<long long> summax[4 * MAXN], sumreg[4 * MAXN];
vector<pair<int, int> > v[4 * MAXN];
set<pair<pair<int, int>, int> > range;
set<pair<pair<int, int>, int> >::iterator it;
void merge(vector<pair<int, int> > &k, vector<pair<int, int> > &l,
           vector<pair<int, int> > &r) {
  int lit = 0, rit = 0;
  k.resize(l.size() + r.size());
  for (int i = 0; i < l.size() + r.size(); ++i)
    if (lit == l.size() || rit == r.size())
      k[i] = (lit == l.size() ? r[rit++] : l[lit++]);
    else if (l[lit] < r[rit])
      k[i] = l[lit++];
    else
      k[i] = r[rit++];
}
void calc(int x, vector<pair<int, int> > &k) {
  summax[x].resize(k.size() + 1);
  sumreg[x].resize(k.size() + 1);
  summax[x][0] = 0;
  sumreg[x][k.size()] = 0;
  for (int i = 1; i < k.size() + 1; ++i)
    summax[x][i] = summax[x][i - 1] + lim[k[i - 1].second];
  for (int i = k.size() - 1; i >= 0; --i)
    sumreg[x][i] = sumreg[x][i + 1] + reg[k[i].second];
}
void build(int x, int l, int r) {
  if (l == r) {
    v[x].push_back(
        make_pair(reg[l] ? (lim[l] + reg[l] - 1) / reg[l] : 1e9 + 9, l));
    calc(x, v[x]);
    return;
  }
  int md = (l + r) / 2;
  build(x * 2, l, md);
  build(x * 2 + 1, md + 1, r);
  merge(v[x], v[x * 2], v[x * 2 + 1]);
  calc(x, v[x]);
}
long long que(int x, int l, int r, int ll, int rr, int time) {
  if (l > rr || r < ll) return 0;
  if (l >= ll && r <= rr) {
    int k = upper_bound(v[x].begin(), v[x].end(), make_pair(time, (int)1e9)) -
            v[x].begin();
    return summax[x][k] + sumreg[x][k] * time;
  }
  int md = (l + r) / 2;
  return que(x * 2, l, md, ll, rr, time) +
         que(x * 2 + 1, md + 1, r, ll, rr, time);
}
void check(set<pair<pair<int, int>, int> >::iterator i, int l, int r) {
  int nl = i->first.second, nr = i->first.first, time = i->second;
  if (nl < l && nr >= l) range.insert(make_pair(make_pair(l - 1, nl), time));
  if (nl <= r && nr > r) range.insert(make_pair(make_pair(nr, r + 1), time));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> s[i] >> lim[i] >> reg[i];
  build(1, 1, n);
  cin >> m;
  range.insert(make_pair(make_pair(n, 1), -1));
  for (int i = 0; i < m; ++i) {
    int t, l, r;
    cin >> t >> l >> r;
    long long ans = 0;
    int lit = l, nl, nr, time;
    while (lit <= r) {
      it = range.lower_bound(make_pair(make_pair(lit, -1), -1));
      check(it, l, r);
      nl = it->first.second;
      nr = it->first.first;
      time = it->second;
      if (time == -1)
        for (; lit <= r && lit <= nr; ++lit)
          ans += min(1LL * lim[lit], 1LL * s[lit] + 1LL * reg[lit] * t);
      else
        ans += que(1, 1, n, lit, min(nr, r), t - time);
      range.erase(it);
      lit = nr + 1;
    }
    range.insert(make_pair(make_pair(r, l), t));
    cout << ans << '\n';
  }
}