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;
long long x[1003], y[1003];
long long dist[1003][1003];
vector<int> g[1003];
int vis[1003];
bool dfs(int u, int k) {
  if (u == n) return true;
  if (vis[u]) return dfs(u + 1, k);
  for (int i = 0; i < g[u].size(); i++) k -= (!vis[g[u][i]]++);
  if (k >= 0 && dfs(u + 1, k)) return true;
  vis[u] = 1;
  int q = k;
  for (int i = 0; i < g[u].size(); i++) k += (!--vis[g[u][i]]);
  if (--k >= 0 && k > q && dfs(u + 1, k)) return true;
  vis[u] = 0;
  return false;
}
bool check(long long h) {
  for (int i = 0; i < n; i++) vis[i] = 0;
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++)
      if (dist[i][j] > h) {
        g[i].push_back(j);
        g[j].push_back(i);
      }
  return dfs(0, k);
}
int main() {
  cin >> n >> k;
  vector<int> d;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) {
      long long nah =
          (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
      dist[i][j] = nah;
      dist[j][i] = nah;
      d.push_back(nah);
    }
  sort(d.begin(), d.end());
  vector<int> fs;
  fs = d;
  d.clear();
  for (int i = 0; i < fs.size(); i++)
    if (i == 0 || fs[i] != fs[i - 1]) d.push_back(fs[i]);
  int l, r, mid;
  l = -1, r = d.size();
  while (l + 1 < r) {
    mid = (r + l) / 2;
    if (check(d[mid]))
      r = mid;
    else
      l = mid;
  }
  check(d[r]);
  for (int i = 0; i < n && k; i++)
    if (vis[i]) {
      cout << i + 1 << " ";
      k--;
    }
  for (int i = 0; i < n && k; i++)
    if (!vis[i]) {
      cout << i + 1 << " ";
      k--;
    }
  return 0;
}