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 maxn = 1003, inf = 2100000000;
int x[maxn], y[maxn], path[maxn], n, K, dist[maxn][maxn], num[maxn * maxn],
    Tnum = 0;
int Link[maxn], pre[maxn * maxn], t[maxn * maxn];
int get() {
  int f = 0, v = 0;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') break;
  if (ch == '-')
    f = 1;
  else
    v = ch - 48;
  while (isdigit(ch = getchar())) v = v * 10 + ch - 48;
  if (f == 1)
    return -v;
  else
    return v;
}
bool dfs(int x, int k) {
  if (x == n + 1) return 1;
  if (path[x] != 0) return dfs(x + 1, k);
  int cnt = 0;
  for (int i = Link[x]; i > 0; i = pre[i])
    if (path[t[i]] == 0) cnt++, path[t[i]] = x;
  if (k >= cnt && dfs(x + 1, k - cnt)) return 1;
  for (int i = Link[x]; i > 0; i = pre[i])
    if (path[t[i]] == x) path[t[i]] = 0;
  if (cnt <= 1) return 0;
  path[x] = x;
  if (k > 0 && dfs(x + 1, k - 1)) return 1;
  path[x] = 0;
  return 0;
}
bool check(int dis) {
  memset(Link, 0, sizeof(Link));
  memset(path, 0, sizeof(path));
  for (int i = 1, tot = 0; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (dist[i][j] > dis) pre[++tot] = Link[i], Link[i] = tot, t[tot] = j;
  return dfs(1, K);
}
int main() {
  n = get(), K = get();
  for (int i = 1; i <= n; i++) x[i] = get(), y[i] = get();
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      num[Tnum++] = dist[i][j] = dist[j][i] =
          (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
  num[Tnum++] = 0, num[Tnum++] = inf;
  sort(num, num + Tnum);
  Tnum = unique(num, num + Tnum) - num;
  int l = 0, r = Tnum - 1;
  while (l < r) {
    int mid = (l + r) / 2;
    if (check(num[mid]))
      r = mid;
    else
      l = mid + 1;
  }
  check(num[l]);
  for (int i = 1; i <= n; i++)
    if (path[i] > 0) printf("%d ", i), K--;
  for (int i = 1; K > 0 && i <= n; i++)
    if (path[i] == 0) printf("%d ", i), K--;
  return 0;
}