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;
int n, m;
bool debug = false;
int k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
int x[1005], y[1005], del[1005], dis[1005][1005];
vector<int> mp[1005];
bool dfs(int x, int k) {
  if (k < 0) return 0;
  if (x >= n) return 1;
  if (del[x]) return dfs(x + 1, k);
  for (int j : mp[x]) {
    if (del[j] == 0) k--;
    del[j]++;
  }
  if (dfs(x + 1, k)) return 1;
  int t = k;
  for (int j : mp[x]) {
    del[j]--;
    if (del[j] == 0) k++;
  }
  del[x] = 1;
  if (k - 1 > t && dfs(x + 1, k - 1)) return 1;
  del[x] = 0;
  return 0;
}
bool ck(int d) {
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    del[i] = 0;
    mp[i].clear();
  }
  for (int(i) = 0; (i) < (int)(n); (i)++)
    for (int(j) = 0; (j) < (int)(n); (j)++)
      if (dis[i][j] > d) {
        mp[i].push_back(j);
      }
  return dfs(0, k);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = 0; (i) < (int)(n); (i)++) {
    scanf("%d%d", x + i, y + i);
  }
  int d = -1;
  for (int(i) = 0; (i) < (int)(n); (i)++)
    for (int(j) = 0; (j) < (int)(n); (j)++) {
      dis[i][j] = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
      d = max(d, dis[i][j]);
    }
  int l = 0, r = d;
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    if (ck(mid))
      r = mid;
    else
      l = mid + 1;
  }
  if (!ck(l)) ck(r);
  for (int(i) = 0; (i) < (int)(n); (i)++)
    if (del[i]) {
      printf("%d ", i + 1);
      k--;
    }
  for (int i = 0; i < n && k > 0; i++)
    if (!del[i]) {
      printf("%d ", i + 1);
      k--;
    }
  return 0;
}