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 = 1000 + 5;
int n, k, vis[maxn];
long long x[maxn], y[maxn], maxd, dd[maxn][maxn];
vector<int> g[maxn];
long long d[maxn * maxn / 2];
long long dis(int a, int b) {
  return (x[a] - x[b]) * (x[a] - x[b]) + (y[a] - y[b]) * (y[a] - y[b]);
}
bool dfs(int cur, int now) {
  if (cur >= n) return true;
  if (g[cur].size() == 0 || vis[cur]) return dfs(cur + 1, now);
  for (int i = 0; i < g[cur].size(); i++) {
    vis[g[cur][i]]++;
    if (vis[g[cur][i]] == 1) now++;
  }
  if (now <= k && dfs(cur + 1, now)) return true;
  int tmp = now;
  for (int i = 0; i < g[cur].size(); i++) {
    vis[g[cur][i]]--;
    if (vis[g[cur][i]] == 0) now--;
  }
  if (g[cur].size() != 1) {
    vis[cur]++;
    if (now <= k && now + 1 < tmp && dfs(cur + 1, now + 1)) return true;
    vis[cur]--;
  }
  return false;
}
bool check(long long ans) {
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (dd[i][j] > ans) {
        g[i].push_back(j);
        g[j].push_back(i);
      }
  return dfs(0, 0);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%I64d %I64d", &x[i], &y[i]);
  int cnt = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      dd[i][j] = dd[j][i] = dis(i, j);
      d[++cnt] = dd[i][j];
    }
  sort(d + 1, d + 1 + cnt);
  cnt = unique(d + 1, d + cnt + 1) - d - 1;
  int l = 1, r = cnt, m;
  while (l < r) {
    m = (l + r) >> 1;
    if (check(d[m]))
      r = m;
    else
      l = m + 1;
  }
  check(d[l]);
  int s = k;
  for (int i = 0; i < n; i++)
    if (vis[i]) printf("%d ", i + 1), s--;
  for (int i = 0; i < n && s > 0; i++)
    if (!vis[i]) printf("%d ", i + 1), s--;
  printf("\n");
  return 0;
}