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 = 100000 + 10;
struct P {
  int m, ra, t;
  inline void get() { t = (ra ? (m - 1) / ra + 1 : 2000000000); }
  bool operator<(const P &rhs) const { return t < rhs.t; }
};
struct node {
  node *l, *r;
  bool ok;
  int last, s;
  vector<int> t;
  vector<long long> ms, ras;
  node(int _s = 0) {
    s = _s;
    last = ok = 0;
    l = r = NULL;
  }
};
void push(node *u) {
  if (!u->ok) return;
  u->l->last = u->last, u->l->ok = 1;
  u->r->last = u->last, u->r->ok = 1;
  u->ok = 0;
}
int s[maxn], ra[maxn], m[maxn];
P p[maxn], tmp[maxn];
node *build(int l, int r) {
  int sz = r - l + 2;
  node *u;
  if (l == r)
    u = new node(s[l]);
  else {
    int mid = (l + r) / 2;
    u = new node();
    u->l = build(l, mid);
    u->r = build(mid + 1, r);
    merge(p + l, p + mid + 1, p + mid + 1, p + r + 1, tmp + l);
    for (int i = l; i <= r; i++) p[i] = tmp[i];
  }
  u->ms.push_back(0);
  u->ras.push_back(0);
  for (int i = l; i <= r; i++)
    u->t.push_back(p[i].t), u->ms.push_back(u->ms.back() + p[i].m),
        u->ras.push_back(u->ras.back() + p[i].ra);
  return u;
}
long long query(int l, int r, int L, int R, node *u, int t) {
  if (L == R) {
    long long ret =
        min((long long)m[L], s[L] + (long long)(t - u->last) * ra[L]);
    u->ok = 1;
    u->last = t;
    s[L] = 0;
    return ret;
  }
  if (l == L && r == R && u->ok) {
    const vector<int> &v = u->t;
    int id = upper_bound(v.begin(), v.end(), t - u->last) - v.begin() - 1;
    long long ret =
        u->ms[id + 1] + (u->ras.back() - u->ras[id + 1]) * (t - u->last);
    u->last = t;
    return ret;
  }
  push(u);
  int mid = (L + R) / 2;
  long long ret;
  if (r <= mid)
    ret = query(l, r, L, mid, u->l, t);
  else if (l > mid)
    ret = query(l, r, mid + 1, R, u->r, t);
  else
    ret =
        query(l, mid, L, mid, u->l, t) + query(mid + 1, r, mid + 1, R, u->r, t);
  if (l == L && r == R) u->ok = 1, u->last = t;
  return ret;
}
int main() {
  srand(time(NULL));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d", &s[i], &m[i], &ra[i]), p[i].m = m[i], p[i].ra = ra[i],
                                           p[i].get();
  node *root = build(1, n);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    printf("%I64d\n", query(l, r, 1, n, root, t));
  }
}