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 N = 1000;
int ans[N], used[N], deg[N], n, k, l, r, q, cur, x[N], y[N], dist[N][N];
vector<int> d, adj[N];
void destroy(int v) {
  used[v]++;
  if (used[v] != 1) return;
  ans[cur++] = v;
  for (int i = 0; i < ((int)(adj[v]).size()); ++i)
    if (!used[adj[v][i]]) {
      deg[adj[v][i]]--;
      deg[v]--;
    }
}
void undestroy(int v) {
  used[v]--;
  if (used[v] != 0) return;
  assert(ans[--cur] == v);
  for (int i = 0; i < ((int)(adj[v]).size()); ++i)
    if (!used[adj[v][i]]) {
      deg[adj[v][i]]++;
      deg[v]++;
    }
}
bool go(int i) {
  if (cur > k) return false;
  if (i == n) {
    for (int j = 0; j < n; ++j)
      if (!used[j]) ans[cur++] = j;
    return true;
  }
  if (deg[i] == 0)
    return go(i + 1);
  else {
    if (deg[i] != 1) {
      destroy(i);
      if (go(i + 1)) return true;
      undestroy(i);
    }
    for (int j = 0; j < ((int)(adj[i]).size()); ++j) destroy(adj[i][j]);
    if (go(i + 1)) return true;
    for (int j = ((int)(adj[i]).size()) - 1; j >= 0; --j) undestroy(adj[i][j]);
    return false;
  }
}
bool check(int limit) {
  for (int i = 0; i < n; ++i) {
    adj[i].clear();
    for (int j = 0; j < n; ++j)
      if (dist[i][j] > limit) adj[i].push_back(j);
    deg[i] = ((int)(adj[i]).size()), used[i] = false;
  }
  cur = 0;
  while (true) {
    bool ok = false;
    for (int i = 0; i < n; ++i)
      if (deg[i] == 1)
        for (int j = 0; j < ((int)(adj[i]).size()); ++j)
          if (!used[adj[i][j]]) {
            destroy(adj[i][j]);
            ok = true;
            break;
          }
    if (!ok) break;
  }
  return (cur > k) ? false : go(0);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d%d", &x[i], &y[i]);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      dist[i][j] =
          ((x[i] - x[j]) * (x[i] - x[j])) + ((y[i] - y[j]) * (y[i] - y[j]));
      d.push_back(dist[i][j]);
    }
  sort(d.begin(), d.end());
  d.resize(unique(d.begin(), d.end()) - d.begin());
  int l = 0, r = ((int)(d).size()) - 1;
  while (l < r) {
    q = (l + r) / 2;
    if (check(d[q]))
      r = q;
    else
      l = q + 1;
  }
  assert(check(d[r]));
  for (int i = 0; i < k; ++i) printf("%d%c", ans[i] + 1, " \n"[i == k - 1]);
  return 0;
}