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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
const string task = "";
template <class T>
T sqr(T first) {
  return first * first;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
long long dist(pair<int, int> a, pair<int, int> b) {
  return sqr(a.first - b.first) + sqr(a.second - b.second);
}
const int nmax = 1010;
pair<int, int> p[nmax];
int n, k, m;
pair<long long, pair<int, int> > e[nmax * nmax];
pair<int, int> die[nmax * 40];
int res[nmax];
int ord[nmax];
vector<int> g[nmax];
int cnt;
int deg[nmax];
set<pair<int, int> > alive;
void kill(int v) {
  res[v] = 1;
  alive.erase(make_pair(-deg[v], v));
  for (int i = 0; i < (int)(g[v].size()); i++) {
    int u = g[v][i];
    if (alive.count(make_pair(-deg[u], u))) {
      alive.erase(make_pair(-deg[u], u));
      deg[u]--;
      if (deg[u] > 0) alive.insert(make_pair(-deg[u], u));
    }
  }
}
void rekill(int v) {
  res[v] = 0;
  for (int i = 0; i < (int)(g[v].size()); i++) {
    int u = g[v][i];
    if (alive.count(make_pair(-deg[u], u)) || deg[u] == 0) {
      alive.erase(make_pair(-deg[u], u));
      deg[u]++;
      alive.insert(make_pair(-deg[u], u));
    }
  }
  alive.insert(make_pair(-deg[v], v));
}
bool go(int kl) {
  if (kl > k) return 0;
  if (alive.size() == 0) return 1;
  pair<int, int> now = *alive.begin();
  int v = now.second;
  if (alive.size() / (-now.first + 1) + kl > k) return 0;
  if (now.first == -1) {
    int w = alive.size();
    assert(w % 2 == 0);
    if (kl + w / 2 > k) return 0;
    while (alive.size() > 0) {
      pair<int, int> now = *alive.begin();
      kill(now.second);
      kl++;
    }
    assert(kl <= k);
    return 1;
  }
  if (kl + deg[v] <= k) {
    vector<int> buf;
    int nkl = kl;
    for (int i = 0; i < (int)(g[v].size()); i++)
      if (res[g[v][i]] == 0) {
        buf.push_back(g[v][i]);
        kill(g[v][i]);
        nkl++;
      }
    if (go(nkl)) return 1;
    for (int i = (int)(buf.size()) - 1; i >= 0; i--) rekill(buf[i]);
  }
  if (kl + 1 <= k) {
    kill(v);
    if (go(kl + 1)) return 1;
    rekill(v);
  }
  return 0;
}
bool check(int st) {
  st++;
  int e1 = 0;
  for (int i = st; i < m; i++) die[e1++] = e[i].second;
  for (int i = 0; i < (int)(n); i++) g[i].clear();
  for (int i = 0; i < (int)(e1); i++) {
    g[die[i].first].push_back(die[i].second);
    g[die[i].second].push_back(die[i].first);
  }
  for (int i = 0; i < (int)(n); i++) deg[i] = g[i].size();
  alive.clear();
  for (int i = 0; i < (int)(n); i++)
    if (deg[i] > 0) alive.insert(make_pair(-deg[i], i));
  memset(res, 0, sizeof res);
  return go(0);
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < (int)(n); i++) scanf("%d %d", &p[i].first, &p[i].second);
  m = 0;
  for (int i = 0; i < (int)(n); i++)
    for (int j = i + 1; j < n; j++)
      e[m++] = make_pair(dist(p[i], p[j]), make_pair(i, j));
  sort(e, e + m);
  int r = m - 1, l = max(0, r - n * k);
  while (l < r) {
    int mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  check(l);
  vector<int> buf;
  for (int i = 0; i < (int)(n); i++)
    if (res[i] > 0)
      res[i] = 1;
    else
      buf.push_back(i);
  while (buf.size() > n - k) {
    res[buf.back()] = 1;
    buf.pop_back();
  }
  for (int i = 0; i < (int)(n); i++)
    if (res[i]) cout << i + 1 << " ";
  cout << endl;
  return 0;
}