Problem: You are given n points on the plane. You need to delete exactly k of them (k < n) so that the diameter of the set of the remaining n - k points were as small as possible. The diameter of a set of points is the maximum pairwise distance between the points of the set. The diameter of a one point set equals zero.

Input

The first input line contains a pair of integers n, k (2 ≤ n ≤ 1000, 1 ≤ k ≤ 30, k < n) — the numbers of points on the plane and the number of points to delete, correspondingly.

Next n lines describe the points, one per line. Each description consists of a pair of integers xi, yi (0 ≤ xi, yi ≤ 32000) — the coordinates of the i-th point. The given points can coincide.

Output

Print k different space-separated integers from 1 to n — the numbers of points to delete. The points are numbered in the order, in which they are given in the input from 1 to n. You can print the numbers in any order. If there are multiple solutions, print any of them.

Examples

Input

5 2
1 2
0 0
2 2
1 1
3 3


Output

5 2

Input

4 1
0 0
0 0
1 1
1 1


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1010;
int n, k;
int x[MaxN], y[MaxN];
int d[MaxN][MaxN];
vector<int> p;
void init() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i];
  p.push_back(0);
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      p.push_back(d[i][j] = (x[i] - x[j]) * (x[i] - x[j]) +
                            (y[i] - y[j]) * (y[i] - y[j]));
  sort(p.begin(), p.end());
  p.erase(unique(p.begin(), p.end()), p.end());
}
vector<int> v[MaxN];
int u[MaxN];
bool DFS(int x, int k) {
  if (x > n) return true;
  if (u[x]) return DFS(x + 1, k);
  for (vector<int>::iterator it = v[x].begin(); it != v[x].end(); ++it)
    if (!u[*it]++) --k;
  if (k >= 0)
    if (DFS(x + 1, k)) return true;
  int tmp = k;
  for (vector<int>::iterator it = v[x].begin(); it != v[x].end(); ++it)
    if (!--u[*it]) ++k;
  ++u[x];
  if (--k >= 0)
    if (k > tmp)
      if (DFS(x + 1, k)) return true;
  --u[x];
  return false;
}
bool check(int dis) {
  for (int i = 1; i <= n; ++i) {
    v[i].clear();
    u[i] = 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (d[i][j] > dis) {
        v[i].push_back(j);
        v[j].push_back(i);
      }
  return DFS(1, k);
}
void work() {
  int L = 0, R = p.size() - 1;
  int ans = R;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (check(p[mid])) {
      ans = mid;
      R = mid - 1;
    } else
      L = mid + 1;
  }
  check(p[ans]);
  for (int i = 1; i <= n; ++i)
    if (u[i]) {
      cout << i << " ";
      --k;
    }
  for (int i = 1; i <= n; ++i)
    if (!u[i] && k) {
      cout << i << " ";
      --k;
    }
}
int main() {
  init();
  work();
  return 0;
}