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, bo[1010], X[1010], Y[1010], d[1010][1010], l, r, mid, i, j, v[1010];
vector<int> e[1010];
bool dfs(int x, int k) {
  if (k < 0) return 0;
  if (x > n) return 1;
  if (v[x]) return dfs(x + 1, k);
  for (int i = 0; i < e[x].size(); ++i)
    if (!v[e[x][i]]++) k--;
  if (dfs(x + 1, k)) return 1;
  int tmp = k;
  for (int i = 0; i < e[x].size(); ++i)
    if (--v[e[x][i]] == 0) k++;
  v[x] = 1;
  if (tmp < k - 1 && dfs(x + 1, k - 1)) return 1;
  v[x] = 0;
  return 0;
}
bool check(int mid) {
  int i, j, id;
  for (i = 1; i <= n; ++i) e[i].clear(), bo[i] = 0, v[i] = 0;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      if (d[i][j] > mid) e[i].push_back(j);
  return dfs(1, k);
}
int main() {
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d%d", &X[i], &Y[i]);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      d[i][j] = (X[i] - X[j]) * (X[i] - X[j]) + (Y[i] - Y[j]) * (Y[i] - Y[j]);
  l = 0;
  r = 2e9 + 1e8;
  while (l <= r) {
    mid = (0ll + l + r) / 2;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  check(r + 1);
  for (i = 1; i <= n; ++i)
    if (v[i]) printf("%d ", i), k--;
  for (i = 1; i <= n; ++i)
    if (!v[i] && k) printf("%d ", i), k--;
}