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;
struct Point {
  int x, y;
} p[1010];
int n, k, cnt, d[1000010], g[1010][1010], vis[1010];
vector<int> Map[1010];
int dis(int i, int j) {
  return (p[i].x - p[j].x) * (p[i].x - p[j].x) +
         (p[i].y - p[j].y) * (p[i].y - p[j].y);
}
bool dfs(int cnt, int m) {
  if (cnt > n) return 1;
  if (vis[cnt]) return dfs(cnt + 1, m);
  int Len = Map[cnt].size();
  for (int(i) = (0); (i) <= (Len - 1); (i)++) {
    m += !vis[Map[cnt][i]];
    vis[Map[cnt][i]]++;
  }
  if (m <= k && dfs(cnt + 1, m)) return 1;
  int temp = m;
  for (int(i) = (0); (i) <= (Len - 1); (i)++) {
    vis[Map[cnt][i]]--;
    m -= !vis[Map[cnt][i]];
  }
  if (Map[cnt].size() != 1) {
    vis[cnt]++;
    if (m + 1 <= k && m + 1 < temp && dfs(cnt + 1, m + 1)) return true;
    vis[cnt]--;
  }
  return false;
}
bool check(int Dis) {
  memset(vis, 0, sizeof vis);
  for (int(i) = (1); (i) <= (n); (i)++) Map[i].clear();
  for (int(i) = (1); (i) <= (n); (i)++)
    for (int(j) = (i + 1); (j) <= (n); (j)++)
      if (g[i][j] > Dis) {
        Map[i].push_back(j);
        Map[j].push_back(i);
      }
  return dfs(1, 0);
}
int Two_Find(int left, int right) {
  while (left < right) {
    int mid = (left + right) >> 1;
    if (check(d[mid]))
      right = mid;
    else
      left = mid + 1;
  }
  return left;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d %d", &p[i].x, &p[i].y);
  for (int(i) = (1); (i) <= (n); (i)++)
    for (int(j) = (1); (j) <= (n); (j)++) d[++cnt] = g[i][j] = dis(i, j);
  sort(d, d + cnt + 1);
  int num = unique(d, d + cnt + 1) - d - 1;
  int mx = Two_Find(0, num);
  check(d[mx]);
  bool first = false;
  for (int i = n; i >= 1; --i)
    if (vis[i]) {
      if (first) putchar(' ');
      printf("%d", i);
      k--;
      first = true;
    }
  for (int i = n; i >= 1 && k > 0; --i)
    if (!vis[i]) {
      if (first) putchar(' ');
      printf("%d", i);
      k--;
      first = true;
    }
  return 0;
}