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;
struct E {
  int to, nxt;
  double d;
} b[2000005];
int fst[1005], T;
void build(int f, int t, double d) {
  b[++T].to = t;
  b[T].d = d;
  b[T].nxt = fst[f];
  fst[f] = T;
}
struct P {
  int x, y;
} p[1005];
double dis[1005][1005];
double D[2000005];
int tot;
double f(int x) { return (double)(x * x); }
int n, k;
double Max;
void init() {
  scanf("%d%d", &n, &k);
  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++) {
      double x = f(p[i].x - p[j].x) + f(p[i].y - p[j].y);
      x = sqrt(x);
      dis[i][j] = x;
      Max = max(Max, x);
      D[++tot] = x;
    }
  sort(D + 1, D + tot + 1);
  D[++tot] = Max + 1.0;
  tot = unique(D + 1, D + tot + 1) - D - 1;
}
int cd[1005];
int use[1005], ans, h;
int q[1005], fro;
int vis[1005][1005];
void dfs(int x, int u) {
  if (u > k || ans) return;
  if (x > fro) {
    bool b = 0;
    if (h) {
      for (int i = 1; i <= n; i++)
        if (use[i]) printf("%d ", i), b = 1;
      if (!b) printf("1");
    }
    ans = 1;
    return;
  }
  int f = q[x];
  if (use[f])
    dfs(x + 1, u);
  else {
    int a = u;
    for (int i = fst[f]; i; i = b[i].nxt) {
      int v = b[i].to;
      if (!use[v]) a++;
      use[v]++;
    }
    dfs(x + 1, a);
    for (int i = fst[f]; i; i = b[i].nxt) use[b[i].to]--;
    if (cd[f] != 1 && u + 1 < a) {
      use[f] = 1;
      dfs(x + 1, u + 1);
      use[f] = 0;
    }
  }
}
bool check(double x) {
  memset(cd, 0, sizeof(cd));
  memset(fst, 0, sizeof(fst));
  memset(use, 0, sizeof(use));
  T = 0;
  fro = 0;
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (dis[i][j] >= x) {
        build(i, j, dis[i][j]);
        build(j, i, dis[i][j]);
        if (!cd[i]) q[++fro] = i;
        if (!cd[j]) q[++fro] = j;
        cd[i]++;
        cd[j]++;
      }
  ans = 0;
  dfs(1, 0);
  return ans;
}
void work() {
  init();
  int l = 1, r = tot;
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (check(D[mid]))
      r = mid;
    else
      l = mid;
  }
  h = 1;
  bool c = check(D[r]);
}
int main() { work(); }