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 = 1011;
struct pt {
  int x, y;
} p[maxn];
int n, k, vt, tot, del, t[maxn], out[maxn], d[maxn], h[maxn];
int l[1000010], v[1000010], x[1000010], dis[maxn][maxn];
int sqr(int x) { return x * x; }
int calc(pt a, pt b) { return sqr(a.x - b.x) + sqr(a.y - b.y); }
void res(int x) {
  if (--t[x] != 0) return;
  --del;
  for (int i = h[x]; i; i = l[i])
    if (!t[v[i]]) {
      ++d[v[i]];
      ++d[x];
    }
}
void rem(int x) {
  if (++t[x] != 1) return;
  ++del;
  for (int i = h[x]; i; i = l[i])
    if (!t[v[i]]) {
      --d[v[i]];
      --d[x];
    }
}
bool dfs(int x) {
  if (del > k) return 0;
  if (x > n) {
    out[0] = 0;
    out[n + 1] = n + 1;
    for (int i = 1; i <= n; ++i)
      if (t[i])
        out[++out[0]] = i;
      else
        out[--out[n + 1]] = i;
    return 1;
  }
  if (d[x] == 0) return dfs(x + 1);
  if (d[x] > 1) {
    rem(x);
    if (dfs(x + 1)) return 1;
    res(x);
  }
  for (int i = h[x]; i; i = l[i]) rem(v[i]);
  if (dfs(x + 1)) return 1;
  for (int i = h[x]; i; i = l[i]) res(v[i]);
  return 0;
}
bool check(int len) {
  del = vt = 0;
  memset(h, 0, sizeof(h));
  memset(d, 0, sizeof(d));
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (dis[i][j] > len) {
        v[++vt] = j;
        l[vt] = h[i];
        h[i] = vt;
        ++d[i];
      }
  return dfs(1);
}
int main() {
  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; j <= n; ++j) {
      dis[i][j] = dis[j][i] = calc(p[i], p[j]);
      x[++tot] = dis[i][j];
    }
  sort(x + 1, x + tot + 1);
  tot = unique(x + 1, x + tot + 1) - x - 1;
  int l = 1, r = tot, mid;
  while (l < r) {
    mid = (l + r) >> 1;
    if (check(x[mid]))
      r = mid;
    else
      l = mid + 1;
  }
  check(x[l]);
  for (int i = 1; i <= k; ++i) printf("%d ", out[i]);
  return 0;
}