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 maxn = 1000 + 15;
int n, K, x[maxn], y[maxn], del[maxn], dist[maxn][maxn], deg[maxn];
vector<int> consider, G[maxn];
bool Dfs(int y, int k, int limit) {
  if (k < 0) return false;
  if (y == consider.size()) return true;
  int x = consider[y];
  if (del[x]) return Dfs(y + 1, k, limit);
  int degree = 0;
  vector<int> fd;
  for (auto&& i : G[x])
    if (!del[i] && dist[x][i] > limit) ++degree, fd.emplace_back(i);
  if (degree == 0) return Dfs(y + 1, k, limit);
  if (degree > 1) {
    del[x] = 1;
    if (Dfs(y + 1, k - 1, limit)) return true;
    del[x] = 0;
  }
  for (auto&& x : fd) del[x] = 1;
  if (Dfs(y + 1, k - (int)fd.size(), limit)) return true;
  for (auto&& x : fd) del[x] = 0;
  return false;
}
bool Check(int limit) {
  consider.clear();
  memset(del, 0, n << 2);
  memset(deg, 0, n << 2);
  for (int i = 0; i < n; ++i) G[i].clear();
  int df = K;
  for (int i = 0; i < n; ++i) {
    int degree = 0;
    for (int j = 0; j < n; ++j)
      if (dist[i][j] > limit && !del[j]) ++degree;
    if (degree > K) {
      del[i] = 1;
      --df;
    }
  }
  for (int i = 0; i < n; ++i)
    if (!del[i]) {
      consider.emplace_back(i);
      for (int j = 0; j < n; ++j)
        if (!del[j] && dist[i][j] > limit) ++deg[i], G[i].push_back(j);
    }
  sort(consider.begin(), consider.end(),
       [&](const int& x, const int& y) { return deg[x] > deg[y]; });
  return Dfs(0, df, limit);
}
int main(int argc, char* argv[]) {
  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]);
  int l = 0, r = 2048000000;
  while (l < r) {
    int mid = l + ((r - l) >> 1);
    if (Check(mid))
      r = mid;
    else
      l = mid + 1;
  }
  Check(l);
  int pf = K;
  for (int i = 0; i < n; ++i)
    if (del[i]) {
      --pf;
      printf("%d ", i + 1);
    }
  for (int i = 0; i < n && pf; ++i)
    if (!del[i]) {
      printf("%d ", i + 1);
      --pf;
    }
  puts("");
  return 0;
}