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 MAX = 1090;
int d[MAX * MAX], dnt;
int dist[MAX][MAX];
struct node {
  int x, y;
};
node ns[MAX];
int n, k;
vector<int> G[MAX];
int vist[MAX];
void build(int limit) {
  for (int i = 1; i <= n; i++) G[i].clear(), vist[i] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (dist[i][j] > limit) {
        G[i].push_back(j), G[j].push_back(i);
      }
    }
  }
}
bool check(int bh, int num) {
  if (bh > n) return 1;
  if (vist[bh]) return check(bh + 1, num);
  if (G[bh].size() > k) {
    vist[bh]++;
    if (num + 1 <= k && check(bh + 1, num + 1)) return 1;
    vist[bh]--;
    return 0;
  } else {
    for (int i = 0; i < (int)G[bh].size(); i++) {
      int u = G[bh][i];
      num += !vist[u];
      vist[u]++;
    }
    if (num <= k && check(bh + 1, num)) return 1;
    int tmp = num;
    for (int i = 0; i < (int)G[bh].size(); i++) {
      int u = G[bh][i];
      vist[u]--;
      num -= !vist[u];
    }
    if (G[bh].size() > 1) {
      vist[bh]++;
      if (num + 1 <= k && num + 1 < tmp && check(bh + 1, num + 1)) return 1;
      vist[bh]--;
    }
    return 0;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &ns[i].x, &ns[i].y);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      d[dnt++] = dist[i][j] = (ns[i].x - ns[j].x) * (ns[i].x - ns[j].x) +
                              (ns[i].y - ns[j].y) * (ns[i].y - ns[j].y);
    }
  }
  sort(d, d + dnt);
  int num = unique(d, d + dnt) - d;
  int l = 0, r = num - 1;
  while (l < r) {
    int mid = (l + r) / 2;
    build(d[mid]);
    bool ok = check(1, 0);
    if (ok)
      r = mid;
    else
      l = mid + 1;
  }
  build(d[l]);
  check(1, 0);
  for (int i = n; i >= 1; i--) {
    if (vist[i]) {
      printf("%d ", i);
      k--;
    }
  }
  for (int i = n; i >= 1 && k > 0; i--) {
    if (!vist[i]) {
      printf("%d ", i);
      k--;
    }
  }
  printf("\n");
  return 0;
}