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 MAX_N = 1000;
int n, k;
int x[MAX_N], y[MAX_N];
int dt[MAX_N][MAX_N];
int dis(int i, int j) {
  return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
}
int ds[MAX_N * MAX_N], nDs;
int bad[MAX_N][MAX_N], nBad[MAX_N];
int cnt[MAX_N], bd;
int perm[MAX_N];
void rec(int it) {
  if (bd > k) {
    return;
  }
  if (it == n) {
    throw 1;
  }
  int i = perm[it];
  if (cnt[i] > 0) {
    rec(it + 1);
    return;
  }
  int newbad = 0;
  for (int j = 0; j < nBad[i]; ++j) {
    if (++cnt[bad[i][j]] == 1) {
      ++bd;
      newbad++;
    }
  }
  rec(it + 1);
  for (int j = 0; j < nBad[i]; ++j) {
    if (--cnt[bad[i][j]] == 0) --bd;
  }
  if (newbad > 1) {
    ++cnt[i];
    ++bd;
    rec(it + 1);
    --cnt[i];
    --bd;
  }
}
int cmp(int a, int b) { return nBad[a] < nBad[b]; }
bool check(int L) {
  for (int i = 0; i < n; ++i) {
    nBad[i] = 0;
    for (int j = 0; j < n; ++j) {
      if (dt[i][j] > L) {
        bad[i][nBad[i]++] = j;
      }
    }
    perm[i] = i;
  }
  sort(perm, perm + n, cmp);
  memset(cnt, 0, sizeof cnt);
  bd = 0;
  try {
    rec(0);
  } catch (int e) {
    return true;
  }
  return false;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      dt[i][j] = dt[j][i] = dis(i, j);
      ds[nDs++] = dt[i][j];
    }
  }
  sort(ds, ds + nDs);
  int l = -1, r = nDs;
  while (l + 1 < r) {
    int m = (l + r) / 2;
    if (check(ds[m]))
      r = m;
    else
      l = m;
  }
  check(ds[r]);
  bool used[MAX_N] = {};
  vector<int> ans;
  for (int i = 0; i < n; ++i) {
    if (cnt[i] > 0) ans.push_back(i + 1), used[i] = true;
  }
  for (int i = 0; i < n; ++i) {
    if (ans.size() < k && !used[i]) {
      used[i] = true;
      ans.push_back(i + 1);
    }
  }
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}