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 p[1000020], c, d[1020][1020];
int n, k, v[1020], x[1020], y[1020];
vector<int> a[1020];
int dfs(int x, int k) {
  if (x == n) return 1;
  if (v[x]) return dfs(x + 1, k);
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    k -= !v[*i]++;
  int u = k;
  if (k >= 0 && dfs(x + 1, k)) return 1;
  for (__typeof(a[x].begin()) i = a[x].begin(); i != a[x].end(); i++)
    k += !--v[*i];
  v[x] = 1;
  if (--k >= 0 && k > u && dfs(x + 1, k)) return 1;
  v[x] = 0;
  return 0;
}
int ok(int x) {
  for (int i = 0; i < n; i++) v[i] = 0, a[i].clear();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (d[i][j] > x) a[i].push_back(j);
  return dfs(0, k);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d%d", x + i, y + i);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      p[c++] = d[i][j] = d[j][i] =
          (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
  p[c++] = 0;
  sort(p, p + c);
  c = unique(p, p + c) - p;
  int l = -1, r = c, m;
  for (; l < r - 1;) m = l + r >> 1, (ok(p[m]) ? r : l) = m;
  ok(p[r]);
  for (int i = 0; i < n; i++)
    if (v[i]) printf("%d ", i + 1), k--;
  for (int i = 0; i < n; i++)
    if (k && !v[i]) printf("%d ", i + 1), k--;
  return 0;
}