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 inf = ~0U >> 1;
const long long INF = ~0LLU >> 1;
int rd() { return RAND_MAX == 32767 ? ((rand() << 15) ^ rand()) : rand(); }
const int maxn = 1010;
int n, M, flag;
int d[maxn], x[maxn], y[maxn], del[maxn], l[maxn][maxn];
vector<int> D, G[maxn];
vector<pair<int, int> > V;
void Dfs(int t, int dep, int E, int m) {
  if (t == dep) {
    flag = 1;
    return;
  }
  int cnt = V[t].second;
  if (d[cnt] * m < E) return;
  if (del[cnt] > 0) {
    Dfs(t + 1, dep, E, m);
    return;
  }
  int sum = 0;
  for (__typeof(G[cnt].begin()) e = G[cnt].begin(); e != G[cnt].end(); e++)
    if (del[*e]++ == 0) {
      sum++;
      E--;
    }
  if (m >= sum) Dfs(t + 1, dep, E, m - sum);
  if (flag) return;
  for (__typeof(G[cnt].begin()) e = G[cnt].begin(); e != G[cnt].end(); e++)
    if (--del[*e] == 0) E++;
  del[cnt]++;
  sum = 0;
  for (__typeof(G[cnt].begin()) e = G[cnt].begin(); e != G[cnt].end(); e++)
    if (del[*e] == 0) sum++;
  if (m >= 1 && sum > 1) Dfs(t + 1, dep, E - sum, m - 1);
  if (flag) return;
  del[cnt]--;
}
int Can(int lim) {
  for (int i = 0; i < n; i++) d[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (l[i][j] > lim) d[i]++;
  int m = M;
  for (int i = 0; i < n; i++) del[i] = 0;
  for (int i = 0; i < n; i++)
    if (d[i] > m) {
      if (m == 0) return 0;
      del[i] = 1;
      d[i] = 0;
      m--;
      for (int j = 0; j < n; j++)
        if (l[i][j] > lim && !del[j]) d[j]--;
    }
  V.clear();
  for (int i = 0; i < n; i++) G[i].clear();
  int e = 0;
  for (int i = 0; i < n; i++)
    if (!del[i] && d[i] > 0) {
      V.push_back(make_pair(d[i], i));
      e++;
      for (int j = 0; j < n; j++)
        if (!del[j] && l[i][j] > lim) G[i].push_back(j);
    }
  e /= 2;
  sort(V.begin(), V.end());
  reverse(V.begin(), V.end());
  flag = 0;
  Dfs(0, V.size(), e, m);
  return flag;
}
int main() {
  cin >> n >> M;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      l[i][j] = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
      D.push_back(l[i][j]);
    }
  sort(D.begin(), D.end());
  D.resize(unique(D.begin(), D.end()) - D.begin());
  int lt, rt, md;
  for (lt = 0, rt = D.size() - 1; lt < rt;) {
    int mid = (lt + rt) / 2;
    Can(D[mid]) ? rt = mid : lt = mid + 1;
  }
  Can(D[lt]);
  int tot = 0;
  for (int i = 0; i < n; i++)
    if (del[i]) {
      cout << i + 1 << ' ';
      tot++;
    }
  for (int i = 0; i < n; i++)
    if (!del[i] && tot < M) {
      tot++;
      cout << i + 1 << ' ';
    }
  cout << endl;
  return 0;
}