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 = 1012;
int n, k, t = 0, tot = 0, ans[N], tim[N], du[N];
int w[N * N];
struct Point {
  int x, y;
} a[N];
inline int Dist(Point a, Point b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
struct Edge {
  int u, v;
  Edge *next;
} e[N * N], *P, *point[N];
inline void Add_Edge(int a, int b) {
  P->u = a;
  P->v = b;
  P->next = point[a];
  point[a] = P++;
  ++du[b];
}
inline void Remove(int now) {
  if (++tim[now] > 1) return;
  ++tot;
  for (Edge *j = point[now]; j; j = j->next)
    if (!tim[j->v]) --du[now], --du[j->v];
}
inline void Resume(int now) {
  if (--tim[now] > 0) return;
  --tot;
  for (Edge *j = point[now]; j; j = j->next)
    if (!tim[j->v]) ++du[now], ++du[j->v];
}
bool Dfs(int now) {
  if (tot > k) return false;
  if (now > n) {
    int l = 0, r = n + 1;
    for (int i = 1; i <= n; ++i)
      if (tim[i])
        ans[++l] = i;
      else
        ans[--r] = i;
    return true;
  }
  if (du[now] == 0) return Dfs(now + 1);
  if (du[now] > 1) {
    Remove(now);
    if (Dfs(now + 1)) return true;
    Resume(now);
  }
  for (Edge *j = point[now]; j; j = j->next) Remove(j->v);
  if (Dfs(now + 1)) return true;
  for (Edge *j = point[now]; j; j = j->next) Resume(j->v);
  return false;
}
bool Check(int key) {
  memset(du, 0, sizeof(du));
  memset(tim, 0, sizeof(tim));
  P = e;
  memset(point, 0, sizeof(point));
  tot = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j < i; ++j)
      if (Dist(a[i], a[j]) > key) Add_Edge(i, j), Add_Edge(j, i);
  return Dfs(1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j < i; ++j) w[++t] = Dist(a[i], a[j]);
  sort(w + 1, w + t + 1);
  int l = 1, r = t;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (Check(w[mid]))
      r = mid - 1;
    else
      l = mid + 1;
  }
  for (int i = 1; i <= k; ++i) printf("%d ", ans[i]);
  puts("");
}