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>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
const long long llINF = 9223372036854775807;
const int INF = 2147483647;
const int maxn = 1e3 + 7;
const int maxm = 2e4 + 7;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;
int n, k;
vector<int> pic[maxn];
int dis[maxn][maxn];
int vis[maxn];
vector<int> d;
int x[maxn], y[maxn];
int min_id = INF;
vector<int> ans;
set<int> s;
bool dfs(int cur, int cnt) {
  if (cur > n) {
    int res = k - cnt;
    for (int i = 1; i <= n; i++) {
      if (vis[i])
        ans.push_back(i);
      else if (res > 0) {
        ans.push_back(i);
        res--;
      }
    }
    return true;
  }
  if (vis[cur]) return dfs(cur + 1, cnt);
  for (int i = 0; i < pic[cur].size(); i++) cnt += (!vis[pic[cur][i]]++);
  if (cnt <= k && dfs(cur + 1, cnt)) return true;
  int tmp = cnt;
  for (int i = 0; i < pic[cur].size(); i++) cnt -= (!--vis[pic[cur][i]]);
  vis[cur] = 1;
  if (cnt + 1 <= k && cnt + 1 < tmp && dfs(cur + 1, cnt + 1)) return true;
  vis[cur] = 0;
  return false;
}
bool check(int id) {
  ans.clear();
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < maxn; i++) pic[i].clear();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (dis[i][j] > d[id]) {
        pic[i].push_back(j);
      }
  if (dfs(1, 0)) {
    min_id = min(min_id, id);
    return true;
  }
  return false;
}
int main(int argc, char const *argv[]) {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      dis[i][j] = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
      s.insert(dis[i][j]);
    }
  for (set<int>::iterator it = s.begin(); it != s.end(); it++) d.push_back(*it);
  int l = 0, r = d.size() - 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      r = mid - 1;
    else
      l = mid + 1;
  }
  check(min_id);
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}