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 mn = 1010;
int x[mn], y[mn], n, m, an, i, j, l;
int e[mn][mn], d[mn], del[mn], lo, mid, hi;
int g[mn][mn], ans[mn], v[mn * mn], tt;
bool sol;
int sqr(int x) { return x * x; }
void rm(int r) {
  if (++del[r] != 1) return;
  int i;
  for (i = 1; i <= g[r][0]; ++i)
    if (!del[g[r][i]]) --d[r], --d[g[r][i]];
  ++an;
}
void rp(int r) {
  if (--del[r] != 0) return;
  int i;
  for (i = 1; i <= g[r][0]; ++i)
    if (!del[g[r][i]]) ++d[g[r][i]], ++d[r];
  --an;
}
bool dfs(int r) {
  if (an > m) return 0;
  if (r > n) return 1;
  if (d[r] == 0) return dfs(r + 1);
  if (d[r] > 1) {
    rm(r);
    if (dfs(r + 1)) return 1;
    rp(r);
  }
  int i;
  for (i = 1; i <= g[r][0]; ++i) rm(g[r][i]);
  if (dfs(r + 1)) return 1;
  for (i = 1; i <= g[r][0]; ++i) rp(g[r][i]);
  return 0;
}
bool solve(int v) {
  for (i = 1; i <= n; ++i) g[i][0] = 0;
  memset(d, 0, sizeof d);
  memset(del, 0, sizeof del);
  for (i = 1; i <= n; ++i)
    for (j = i + 1; j <= n; ++j)
      if (e[i][j] > v) {
        g[i][++g[i][0]] = j;
        g[j][++g[j][0]] = i;
        ++d[i], ++d[j];
      }
  an = 0;
  return dfs(1);
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) scanf("%d%d", x + i, y + i);
  v[1] = 0, tt = 1;
  for (i = 1; i <= n; ++i)
    for (j = i + 1; j <= n; ++j) {
      e[i][j] = sqr(x[i] - x[j]) + sqr(y[i] - y[j]);
      v[++tt] = e[i][j];
    }
  sort(v + 1, v + 1 + tt);
  tt = unique(v + 1, v + 1 + tt) - v - 1;
  lo = 1, hi = tt;
  while (lo < hi) {
    mid = (lo + hi) / 2;
    if (solve(v[mid]))
      hi = mid;
    else
      lo = mid + 1;
  }
  solve(v[lo]);
  for (i = 1; i <= n; ++i)
    if (del[i]) ans[++l] = i;
  for (i = 1; i <= n; ++i) {
    if (l == m) break;
    if (!del[i]) ans[++l] = i;
  }
  printf("%d", ans[1]);
  for (i = 2; i <= l; ++i) printf(" %d", ans[i]);
  printf("\n");
  return 0;
}