Today the North Pole hosts an Olympiad in a sport called... toy igloo skyscrapers' building!

There are n walruses taking part in the contest. Each walrus is given a unique number from 1 to n. After start each walrus begins to build his own igloo skyscraper. Initially, at the moment of time equal to 0, the height of the skyscraper i-th walrus is equal to ai. Each minute the i-th walrus finishes building bi floors.

The journalists that are reporting from the spot where the Olympiad is taking place, make q queries to the organizers. Each query is characterized by a group of three numbers li, ri, ti. The organizers respond to each query with a number x, such that:

1. Number x lies on the interval from li to ri inclusive (li ≤ x ≤ ri).

2. The skyscraper of the walrus number x possesses the maximum height among the skyscrapers of all walruses from the interval [li, ri] at the moment of time ti.

For each journalists' query print the number of the walrus x that meets the above-given criteria. If there are several possible answers, print any of them.

Input

The first line contains numbers n and q (1 ≤ n, q ≤ 105). Next n lines contain pairs of numbers ai, bi (1 ≤ ai, bi ≤ 109). Then follow q queries i the following format li, ri, ti, one per each line (1 ≤ li ≤ ri ≤ n, 0 ≤ ti ≤ 106). All input numbers are integers.

Output

For each journalists' query print the number of the walrus x that meets the criteria, given in the statement. Print one number per line.

Examples

Input

5 4
4 1
3 5
6 2
3 5
6 5
1 5 2
1 3 5
1 1 0
1 5 0


Output

5
2
1
5


Input

5 4
6 1
5 1
2 5
4 3
6 1
2 4 1
3 4 5
1 4 5
1 2 0


Output

3
3
3
1
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
constexpr ll inf = 1e18;
struct line {
  ll k, b;
  ld x0;
  int ind;
  line() : k(0), b(inf), x0(-inf), ind(-1){};
  line(ll k, ll b, int ind) : k(k), b(b), ind(ind), x0(-inf){};
  line(ll x0) : k(0), b(inf), x0(x0), ind(-1){};
  ll f(ll x) { return k * x + b; }
};
bool operator<(const line &a, const line &b) { return a.b < b.b; }
ld intersection(const line &a, const line &b) {
  return ld(a.b - b.b) / ld(b.k - a.k);
}
struct cht {
  vector<line> ch;
  void add_line(line l) {
    if (!ch.empty() && ch.back().k <= l.k) return;
    while (!ch.empty() && intersection(ch.back(), l) < ch.back().x0)
      ch.pop_back();
    if (!ch.empty()) l.x0 = intersection(l, ch.back());
    ch.push_back(l);
  }
  pair<ll, int> get_ans(ll x) {
    int l = 0, r = ch.size();
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (ch[m].x0 < x)
        l = m;
      else
        r = m;
    }
    return make_pair(ch[l].f(x), ch[l].ind);
  }
};
vector<cht> t;
void build(const vector<line> &arr, int v, int l, int r) {
  if (r - l == 1)
    t[v].add_line(arr[l]);
  else {
    vector<line> buf(arr.begin() + l, arr.begin() + r);
    sort(buf.begin(), buf.end());
    for (const line &f : buf) t[v].add_line(f);
    int m = (l + r) >> 1;
    build(arr, 2 * v + 1, l, m);
    build(arr, 2 * v + 2, m, r);
  }
}
pair<ll, int> get(int v, int l, int r, int ql, int qr, ll x) {
  if (l >= qr || r <= ql) return make_pair(inf, -1);
  if (l >= ql && r <= qr) return t[v].get_ans(x);
  int m = (l + r) >> 1;
  return min(get(2 * v + 1, l, m, ql, qr, x), get(2 * v + 2, m, r, ql, qr, x));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n = 0, q = 0;
  cin >> n >> q;
  vector<line> arr(n);
  for (int i = 0; i < n; ++i) {
    ll a = 0, b = 0;
    cin >> a >> b;
    arr[i] = line(-b, -a, i);
  }
  t.resize(4 * n);
  build(arr, 0, 0, n);
  while (q--) {
    int l = 0, r = 0, t = 0;
    cin >> l >> r >> t;
    cout << get(0, 0, n, l - 1, r, t).second + 1 << '\n';
  }
  return 0;
}
