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;
int n, k;
struct node {
  int p, q;
} a[1005];
int vis[1005];
int p[1000005];
int r;
vector<int> v[1005];
int ll, rr;
int ans;
int tot;
struct bian {
  int x, y, dist;
} b[1000005];
int tt;
int dis(int i, int j) {
  return (a[i].p - a[j].p) * (a[i].p - a[j].p) +
         (a[i].q - a[j].q) * (a[i].q - a[j].q);
}
void add(int x, int y) { v[x].push_back(y); }
bool dfs(int x, int num) {
  if (x == n + 1) return 1;
  if (vis[x]) return dfs(x + 1, num);
  for (int i = 0; i < (int)v[x].size(); i++) {
    num -= !vis[v[x][i]]++;
  }
  int u = num;
  if (num >= 0 && dfs(x + 1, num)) return 1;
  for (int i = 0; i < (int)v[x].size(); i++) num += !--vis[v[x][i]];
  vis[x] = 1;
  if (--num >= 0 && num > u && dfs(x + 1, num)) return 1;
  vis[x] = 0;
  return 0;
}
bool cmp(bian a, bian b) { return a.dist > b.dist; }
bool judge(int mid) {
  for (int i = 1; i <= n; i++) {
    v[i].clear();
    vis[i] = 0;
  }
  int pp = 0;
  for (int i = 1; i <= tt; i++) {
    if (b[i].dist <= mid) break;
    int x = b[i].x, y = b[i].y;
    add(x, y);
    add(y, x);
  }
  return dfs(1, k);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].p, &a[i].q);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      b[++tt].x = i;
      b[tt].y = j;
      b[tt].dist = dis(i, j);
      p[++r] = dis(i, j);
    }
  }
  p[0] = 0;
  r++;
  sort(p, p + r);
  r = unique(p, p + r) - p;
  sort(b + 1, b + tt + 1, cmp);
  int ll = -1, rr = r;
  while (ll < rr - 1) {
    int mid = (ll + rr) / 2;
    if (judge(p[mid])) {
      rr = mid;
    } else {
      ll = mid;
    }
  }
  judge(p[rr]);
  for (int i = 1; i <= n; i++) {
    if (vis[i]) {
      printf("%d ", i);
      k--;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i] && k) {
      printf("%d ", i);
      k--;
    }
  }
  puts("");
  return 0;
}