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 N = 1005;
int n, k, x[N], y[N];
int D[N * N], dis[N][N];
int cnt, vis[N];
vector<int> e[N];
bool dfs(int x, int cnt) {
  if (x > n) return 1;
  if (vis[x]) return dfs(x + 1, cnt);
  for (auto i : e[x]) cnt -= !(vis[i]++);
  if (cnt >= 0 && dfs(x + 1, cnt)) return 1;
  vis[x] = 1;
  int t = cnt;
  for (auto i : e[x]) cnt += (!--vis[i]);
  if ((--cnt) >= 0 && cnt > t && dfs(x + 1, cnt)) return 1;
  vis[x] = 0;
  return 0;
}
bool check(int mid) {
  for (int i = (1); i <= (int)n; i++) vis[i] = 0, e[i].resize(0);
  for (int i = (1); i <= (int)n; i++)
    for (int j = (1); j <= (int)n; j++)
      if (dis[i][j] > mid) e[i].push_back(j);
  return dfs(1, k);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (int)n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = (1); i <= (int)n; i++)
    for (int j = (i + 1); j <= (int)n; j++)
      D[++*D] = dis[i][j] = ((1ll * (x[i] - x[j]) * (x[i] - x[j])) +
                             (1ll * (y[i] - y[j]) * (y[i] - y[j])));
  sort(D + 1, D + *D + 1);
  D[++*D] = 2147480000;
  int l = 1, r = *D, ans;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(D[mid]))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  check(D[ans]);
  for (int i = (1); i <= (int)n; i++)
    if (vis[i]) printf("%d ", i), k--;
  for (int i = (1); i <= (int)n; i++)
    if (!vis[i] && k) printf("%d ", i), k--;
}