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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
vector<pair<pair<int, long long>, long long> > v[101010 << 2];
int flag[101010 << 2], s[101010], mx[101010], r[101010];
void build(int u, int L, int R) {
  for (int i = L; i < R; i++)
    v[u].push_back(pair<pair<int, long long>, long long>(
        pair<int, long long>(r[i] ? mx[i] / r[i] : 0x3f3f3f3f, r[i]), mx[i]));
  sort(v[u].begin(), v[u].end());
  for (int i = 1; i < v[u].size(); i++) {
    v[u][i].first.second += v[u][i - 1].first.second;
    v[u][i].second += v[u][i - 1].second;
  }
  if (L + 1 == R) return;
  build((u << 1), L, (L + R >> 1));
  build(((u << 1) | 1), (L + R >> 1), R);
}
long long calc(int u, int L, int R, int st, int ed, int T) {
  if (R <= st || ed <= L) return 0;
  if (st <= L && R <= ed && flag[u]) {
    long long ans = v[u].back().first.second * (T - flag[u]);
    int k = lower_bound(v[u].begin(), v[u].end(),
                        pair<pair<int, long long>, long long>(
                            pair<int, long long>(T - flag[u], -1), -1)) -
            v[u].begin();
    if (k) ans += v[u][k - 1].second - v[u][k - 1].first.second * (T - flag[u]);
    flag[u] = T;
    return ans;
  }
  if (L + 1 == R) {
    long long ans = min(v[u][0].first.second * T + s[L], v[u][0].second);
    flag[u] = T;
    s[L] = 0;
    return ans;
  }
  if (flag[u]) {
    if (flag[(u << 1)] < flag[u]) flag[(u << 1)] = flag[u];
    if (flag[((u << 1) | 1)] < flag[u]) flag[((u << 1) | 1)] = flag[u];
  }
  long long ans = calc((u << 1), L, (L + R >> 1), st, ed, T) +
                  calc(((u << 1) | 1), (L + R >> 1), R, st, ed, T);
  flag[u] = flag[(u << 1)] == flag[((u << 1) | 1)] ? flag[(u << 1)] : 0;
  return ans;
}
int main() {
  int n;
  gn(n);
  for (int i = 1; i <= n; i++) gn(s[i], mx[i], r[i]);
  build(1, 1, n + 1);
  int m;
  for (gn(m); m--;) {
    int t, l, r;
    gn(t, l, r);
    println(calc(1, 1, n + 1, l, r + 1, t));
  }
  return 0;
}