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 = 1005;
const int maxm = 500005;
int x[maxn], y[maxn];
int dis[maxm];
vector<int> adj[maxn];
int n, m, K;
void init() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
}
inline int dist2(int i, int j) {
  return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
}
int vis[maxn];
bool dfs(int t, int cnt) {
  if (t > n) return 1;
  if (vis[t]) return dfs(t + 1, cnt);
  int ts = adj[t].size();
  for (int i = 0; i < ts; i++) cnt -= !vis[adj[t][i]], vis[adj[t][i]]++;
  if (cnt >= 0 && dfs(t + 1, cnt)) return 1;
  int tmp = cnt;
  for (int i = 0; i < ts; i++) vis[adj[t][i]]--, cnt += !vis[adj[t][i]];
  cnt--;
  vis[t]++;
  if (cnt >= 0 && cnt > tmp && dfs(t + 1, cnt)) return 1;
  vis[t]--;
  return 0;
}
bool check(int lim) {
  for (int i = 1; i <= n; i++) adj[i].clear(), vis[i] = 0;
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (dist2(i, j) > lim) adj[i].push_back(j), adj[j].push_back(i);
  return dfs(1, K);
}
void work() {
  m = 0;
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) dis[++m] = dist2(i, j);
  sort(dis + 1, dis + m + 1);
  int m0 = m;
  m = 1;
  for (int i = 2; i <= m0; i++)
    if (dis[i] != dis[m]) dis[++m] = dis[i];
  int l = 0, r = m;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(dis[mid]))
      r = mid;
    else
      l = mid + 1;
  }
  check(dis[l]);
}
void print() {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (vis[i]) cnt++, printf("%d ", i);
  for (int i = 1; i <= n && cnt < K; i++)
    if (!vis[i]) cnt++, printf("%d ", i);
  puts("");
}
int main() {
  init();
  work();
  print();
  return 0;
}