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;
vector<int> a[1005];
int d[1005][1005];
int e[1000005];
int h[1005];
int x[1005];
int y[1005];
int n, m, k;
bool dfs(int x, int k) {
  if (x > n) return 1;
  if (h[x]) return dfs(x + 1, k);
  int i, t;
  for (i = 0; i < a[x].size(); ++i) {
    if (h[a[x][i]] == 0) --k;
    ++h[a[x][i]];
  }
  if (k >= 0 && dfs(x + 1, k)) return 1;
  t = max(-1, k);
  for (i = 0; i < a[x].size(); ++i) {
    if (h[a[x][i]] == 1) ++k;
    --h[a[x][i]];
  }
  h[x] = 1, --k;
  if (k > t && dfs(x + 1, k)) return 1;
  h[x] = 0;
  return 0;
}
bool check(int dist) {
  int i, j;
  for (i = 1; i <= n; ++i) {
    a[i].clear();
    h[i] = 0;
  }
  for (i = 1; i < n; ++i)
    for (j = 1; j <= n; ++j)
      if (d[i][j] > dist) {
        a[i].push_back(j);
        a[j].push_back(i);
      }
  return dfs(1, k);
}
void pre() {
  int i, j;
  for (i = 1; i < n; ++i)
    for (j = i + 1; j <= n; ++j)
      e[++m] = d[i][j] =
          (((x[i] - x[j]) * (x[i] - x[j])) + ((y[i] - y[j]) * (y[i] - y[j])));
  sort(&e[1], &e[m + 1]);
  m = unique(&e[1], &e[m + 1]) - e - 1;
}
void work() {
  int i, l, r, mid;
  l = 0, r = m;
  while (l < r) {
    mid = (l + r) / 2;
    if (check(e[mid]))
      r = mid;
    else
      l = mid + 1;
  }
  check(e[l]);
  for (i = 1; i <= n; ++i)
    if (h[i]) {
      printf("%d ", i);
      --k;
    }
  for (i = 1; i <= n; ++i)
    if (k != 0 && h[i] == 0) {
      printf("%d ", i);
      --k;
    }
  printf("\n");
}
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d%d", &x[i], &y[i]);
  pre();
  work();
  return 0;
}