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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = 1005;
vector<int> a[maxn];
int n, k;
int x[maxn], y[maxn], dist[maxn * maxn];
int take[maxn];
bool dfs(int v, int k) {
  if (v == n) return k >= 0;
  if (take[v]) return dfs(v + 1, k);
  int cnt = 0;
  for (int i = 0; i < (int)(a[v]).size(); i++) {
    if (take[a[v][i]] == 0) {
      k--;
      cnt++;
    }
    take[a[v][i]]++;
  }
  if (k >= 0 && dfs(v + 1, k)) return true;
  for (int i = 0; i < (int)(a[v]).size(); i++) {
    if (take[a[v][i]] == 1) {
      k++;
    }
    take[a[v][i]]--;
  }
  take[v]++;
  if (k - 1 >= 0 && cnt > 1 && dfs(v + 1, k - 1)) return true;
  take[v]--;
  return false;
}
bool check(int m) {
  for (int i = 0; i < n; i++) {
    take[i] = 0;
    a[i].clear();
  }
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) > m) {
        a[i].push_back(j);
        a[j].push_back(i);
      }
    }
  }
  return dfs(0, k);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
  }
  int e = 0;
  dist[e++] = 0;
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      dist[e++] = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
    }
  }
  sort(dist, dist + e);
  e = unique(dist, dist + e) - dist;
  int l = -1;
  int r = e - 1;
  while (l < r - 1) {
    int m = (l + r) >> 1;
    if (check(dist[m])) {
      r = m;
    } else {
      l = m;
    }
  }
  check(dist[r]);
  for (int i = 0; i < n; i++) {
    if (take[i]) {
      printf("%d ", i + 1);
      k--;
    }
  }
  for (int i = 0; i < n; i++) {
    if (!take[i] && k > 0) {
      printf("%d ", i + 1);
      k--;
    }
  }
  return 0;
}