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 = 1005;
int n, K, x[N], y[N], choose[N], deg[N];
vector<int> G[N];
vector<int> res;
struct cmp {
  bool operator()(const int& k1, const int& k2) const {
    return deg[k1] > deg[k2];
  }
};
set<int, cmp> S;
void remove(int k1) {
  assert(!choose[k1]);
  choose[k1] = 1;
  for (auto x : G[k1])
    if (!choose[x]) {
      --deg[k1];
      --deg[x];
    }
}
void restore(int k1) {
  assert(choose[k1]);
  choose[k1] = 0;
  for (auto x : G[k1])
    if (!choose[x]) {
      ++deg[k1];
      ++deg[x];
    }
}
bool dfs(int remain) {
  int tt = 0;
  for (int i = (1); i <= (n); ++i) {
    if (choose[i]) ++tt;
  }
  assert(tt + remain == K);
  if (remain == 0) {
    for (int i = (1); i <= (n); ++i)
      if (deg[i]) return 0;
    return 1;
  }
  int pos = -1;
  for (int i = (1); i <= (n); ++i)
    if (!choose[i] && deg[i] == 1) {
      pos = i;
      break;
    }
  if (pos != -1) {
    int tem = -1;
    for (auto x : G[pos])
      if (!choose[x]) {
        tem = x;
        break;
      }
    assert(tem != -1);
    remove(tem);
    if (dfs(remain - 1)) return 1;
    restore(tem);
    return 0;
  }
  int u = 0;
  for (int i = (1); i <= (n); ++i)
    if (!choose[i] && deg[i] > deg[u]) u = i;
  if (deg[u] == 0) {
    remove(u);
    assert(dfs(remain - 1));
    return 1;
  }
  remove(u);
  if (dfs(remain - 1)) return 1;
  restore(u);
  if (deg[u] > 1 && remain - deg[u] >= 0) {
    vector<int> v;
    for (auto x : G[u])
      if (!choose[x]) v.push_back(x);
    assert(((int)(v).size()) == deg[u]);
    int tmp = deg[u];
    for (auto x : v) remove(x);
    if (dfs(remain - tmp)) {
      return 1;
    }
    for (auto x : v) restore(x);
  }
  return 0;
}
bool chk(int mid) {
  for (int i = (1); i <= (n); ++i) G[i].clear(), choose[i] = 0, deg[i] = 0;
  for (int i = (1); i <= (n); ++i)
    for (int j = (i + 1); j <= (n); ++j) {
      if ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) > mid) {
        G[i].push_back(j), G[j].push_back(i);
        ++deg[i], ++deg[j];
      }
    }
  return dfs(K);
}
int main() {
  deg[0] = -1;
  scanf("%d%d", &n, &K);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d%d", &x[i], &y[i]);
  }
  int l = 0, r = 2 * 32000 * 32000, ans = 0;
  while (l <= r) {
    int mid = ((r - l) >> 1) + l;
    ((void)0);
    if (chk(mid))
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  chk(ans);
  ((void)0);
  for (int i = (1); i <= (n); ++i)
    if (choose[i]) printf("%d ", i);
  puts("");
  return 0;
}