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, k;
int x[1010], y[1010];
int dist[1010][1010];
int del[1010];
vector<int> g[1010];
bool dfs(int u, int k) {
  if (u > n) return 1;
  if (del[u]) return dfs(u + 1, k);
  int tmp;
  for (int i = 0; i < g[u].size(); i++)
    if (!del[g[u][i]]++) k--;
  if (k >= 0 && dfs(u + 1, k)) return 1;
  tmp = max(-1, k);
  for (int i = 0; i < g[u].size(); i++)
    if (--del[g[u][i]] == 0) k++;
  del[u] = 1;
  k--;
  if (k > tmp && dfs(u + 1, k)) return 1;
  del[u] = 0;
  return 0;
}
bool ok(int mid) {
  memset(del, 0, sizeof(del));
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (dist[i][j] > mid) g[i].push_back(j), g[j].push_back(i);
  return dfs(1, k);
}
int d[1010 * 1010];
int m;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      d[++m] = dist[i][j] =
          (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
  sort(d, d + m + 1);
  m = unique(d, d + m + 1) - d - 1;
  int lo = 0, hi = m;
  int mid;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (ok(d[mid]))
      hi = mid;
    else
      lo = mid + 1;
  }
  ok(d[lo]);
  for (int i = 1; i <= n; i++)
    if (del[i]) {
      cout << i << ' ';
      k--;
    }
  for (int i = 1; i <= n; i++)
    if (!del[i] && k) {
      cout << i << ' ';
      k--;
    }
  cout << endl;
  return 0;
}