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 = 0x3f3f3f3f;
const double esp = 0.0000000001;
const double PI = acos(-1.0);
const int mod = 1e9 + 7;
const int MY = 1400 + 5;
const int MX = 1010 + 5;
int num, n, k;
vector<int> G[MX];
int d[MX * MX], g[MX][MX], vis[MX];
struct node {
  int x, y;
} P[MX];
bool dfs(int cnt, int m) {
  if (cnt > n) return true;
  if (vis[cnt]) return dfs(cnt + 1, m);
  for (int i = 0; i < (int)G[cnt].size(); ++i) {
    m += !vis[G[cnt][i]];
    vis[G[cnt][i]]++;
  }
  if (m <= k && dfs(cnt + 1, m)) return true;
  int temp = m;
  for (int i = 0; i < (int)G[cnt].size(); ++i) {
    vis[G[cnt][i]]--;
    m -= !vis[G[cnt][i]];
  }
  if (G[cnt].size() != 1) {
    vis[cnt]++;
    if (m + 1 <= k && m + 1 < temp && dfs(cnt + 1, m + 1)) return true;
    vis[cnt]--;
  }
  return false;
}
bool judge(int dist) {
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; ++i) G[i].clear();
  for (int i = 1; i < n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (g[i][j] > dist) {
        G[i].push_back(j);
        G[j].push_back(i);
      }
  return dfs(1, 0);
}
int binary_search(int le, int rt) {
  int mid;
  while (le < rt) {
    mid = (le + rt) >> 1;
    if (judge(d[mid]))
      rt = mid;
    else
      le = mid + 1;
  }
  return le;
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    num = 0;
    d[0] = 0;
    for (int i = 1; i <= n; ++i) scanf("%d%d", &P[i].x, &P[i].y);
    for (int i = 1; i < n; ++i)
      for (int j = i + 1; j <= n; ++j)
        d[++num] = g[i][j] = (P[i].x - P[j].x) * (P[i].x - P[j].x) +
                             (P[i].y - P[j].y) * (P[i].y - P[j].y);
    sort(d, d + num + 1);
    num = unique(d, d + num + 1) - d - 1;
    int mx = binary_search(0, num);
    judge(d[mx]);
    bool first = false;
    for (int i = n; i >= 1; --i)
      if (vis[i]) {
        if (first) putchar(' ');
        printf("%d", i);
        k--;
        first = true;
      }
    for (int i = n; i >= 1 && k > 0; --i)
      if (!vis[i]) {
        if (first) putchar(' ');
        printf("%d", i);
        k--;
        first = true;
      }
    cout << endl;
  }
  return 0;
}