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;
int n, rt[100005], s[100005], m[100005], r[100005], et[100005];
int lc[100005 * 30], rc[100005 * 30], tot;
long long sm[100005 * 30], sr[100005 * 30];
void ins(int &u, int l, int r, int p, int m, int R) {
  sm[++tot] = sm[u] + m;
  sr[tot] = sr[u] + R;
  lc[tot] = lc[u];
  rc[tot] = rc[u];
  u = tot;
  if (l == r) return;
  int M = l + r >> 1;
  p <= M ? ins(lc[u], l, M, p, m, R) : ins(rc[u], M + 1, r, p, m, R);
}
long long qry(int u, int v, int l, int r, int p) {
  if (!u && !v) return 0;
  if (r <= p) return sm[u] - sm[v];
  if (l > p) return (sr[u] - sr[v]) * p;
  int m = l + r >> 1;
  return qry(lc[u], lc[v], l, m, p) + qry(rc[u], rc[v], m + 1, r, p);
}
struct node {
  int l, r;
  mutable int t;
  node(const int &u = 0, const int &v = 0, const int &t = 0)
      : l(u), r(v), t(t) {}
  bool operator<(const node &B) const { return l < B.l; }
};
set<node> st;
set<node>::iterator split(int t) {
  set<node>::iterator it = st.upper_bound(node(t));
  it--;
  if ((*it).l == t) return it;
  int l = (*it).l, r = (*it).r, v = (*it).t;
  st.erase(it);
  st.insert(node(l, t - 1, v));
  return st.insert(node(t, r, v)).first;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &s[i], &m[i], &r[i]);
    rt[i] = rt[i - 1];
    if (r[i]) {
      et[i] = (m[i] + r[i] - 1) / r[i],
      ins(rt[i], 0, (100000), et[i], m[i], r[i]);
    }
    st.insert(node(i, i, -1));
  }
  int q;
  scanf("%d", &q);
  for (int t, l, r; q--;) {
    scanf("%d%d%d", &t, &l, &r);
    set<node>::iterator tr = split(r + 1);
    set<node>::iterator tl = split(l);
    long long ans = 0;
    for (; tl != tr;) {
      if ((*tl).t == -1)
        ans += min(s[(*tl).l] + ::r[(*tl).l] * 1ll * t, 1ll * m[(*tl).l]);
      else
        ans += qry(rt[(*tl).r], rt[(*tl).l - 1], 0, (100000), t - (*tl).t);
      st.erase(tl++);
    }
    st.insert(node(l, r, t));
    printf("%lld\n", ans);
  }
}