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, m, tot, l, r, edgetot, x;
int dis[N * N], point[N], deg[N], can[N], f[N];
struct pp {
  int x, y;
} a[N];
struct edge {
  int v, next;
} e[N * N];
int calc(pp a, pp b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
inline void add_edge(int x, int y) {
  ++edgetot;
  e[edgetot].v = y;
  e[edgetot].next = point[x];
  point[x] = edgetot;
  ++deg[y];
}
inline void remove(int now) {
  if (++can[now] > 1) return;
  ++tot;
  for (int j = point[now]; j; j = e[j].next)
    if (!can[e[j].v]) --deg[now], --deg[e[j].v];
}
inline void resume(int now) {
  if (--can[now] > 0) return;
  --tot;
  for (int j = point[now]; j; j = e[j].next)
    if (!can[e[j].v]) ++deg[now], ++deg[e[j].v];
}
bool dfs(int now) {
  if (tot > m) return false;
  if (now > n) {
    int o = 0;
    for (int i = 1; i <= n; ++i)
      if (can[i]) f[++o] = i;
    if (o < m)
      for (int i = 1; i <= n; ++i)
        if (can[i] == 0) {
          f[++o] = i;
          if (o == m) break;
        }
    return true;
  }
  if (deg[now] == 0) return dfs(now + 1);
  if (deg[now] > 1) {
    remove(now);
    if (dfs(now + 1)) return true;
    resume(now);
  }
  for (int j = point[now]; j; j = e[j].next) remove(e[j].v);
  if (dfs(now + 1)) return true;
  for (int j = point[now]; j; j = e[j].next) resume(e[j].v);
  return false;
}
inline bool check(int x) {
  edgetot = 0;
  tot = 0;
  memset(point, 0, sizeof(point));
  memset(can, 0, sizeof(can));
  memset(deg, 0, sizeof(deg));
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j)
      if (calc(a[i], a[j]) > x) add_edge(i, j), add_edge(j, i);
  return dfs(1);
}
int main() {
  scanf("%d%d", &n, &m);
  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 = i; j <= n; ++j) dis[++tot] = calc(a[i], a[j]);
  sort(dis + 1, dis + tot + 1);
  tot = unique(dis + 1, dis + tot + 1) - dis - 1;
  int l = 1, r = tot;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(dis[mid]))
      r = mid - 1;
    else
      l = mid + 1;
  }
  for (int i = 1; i <= m; ++i) printf("%d ", f[i]);
}