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
#include <bits/stdc++.h>
using namespace std;
class Edge {
 public:
  int v, u, d;
  Edge(int _v, int _u, int _d) : v(_v), u(_u), d(_d) {}
  const bool operator<(const Edge &b) const { return d < b.d; }
};
class Coor {
 public:
  int x, y;
  Coor() {}
  Coor(int _x, int _y) : x(_x), y(_y) {}
  Coor &operator-=(const Coor &b) {
    x -= b.x;
    y -= b.y;
    return *this;
  }
  Coor operator-(const Coor &b) const { return Coor(*this) -= b; }
  const bool operator==(const Coor &b) const { return x == b.x && y == b.y; }
  void input() { scanf("%d %d", &x, &y); }
};
int n, k;
Coor pt[1050];
int src, sink;
vector<int> adj[1050];
int res[1050][1050];
int visited[1050], visid = 0;
int forbidden[1050], fid = 0;
int involve[1050];
int failcnt[1050];
inline int sqr(int x) { return x * x; }
inline int dist2(Coor a, Coor b) { return sqr(a.x - b.x) + sqr(a.y - b.y); }
inline int dist2(int a, int b) { return dist2(pt[a], pt[b]); }
inline int cross(Coor a, Coor b) { return a.x * b.y - a.y * b.x; }
inline int cross(Coor o, Coor a, Coor b) {
  return cross(Coor(a.x - o.x, a.y - o.y), Coor(b.x - o.x, b.y - o.y));
}
inline void append(int v, int u) {
  adj[v].push_back(u);
  adj[u].push_back(v);
  res[v][u] = 1;
  res[u][v] = 0;
}
bool dfs(int v) {
  visited[v] = visid;
  if (v == sink) return 1;
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (visited[u] == visid) continue;
    if (forbidden[u] == fid) continue;
    if (res[v][u] <= 0) continue;
    if (dfs(u)) {
      res[v][u]--;
      res[u][v]++;
      return 1;
    }
  }
  return 0;
}
inline int flow(int maxallow) {
  int f = 0;
  while (1) {
    if (f > maxallow) return f;
    ++visid;
    if (dfs(src))
      f++;
    else
      return f;
  }
}
inline int test(int v, int u) {
  int d2 = dist2(v, u), cnt = 0;
  src = n;
  sink = n + 1;
  adj[src].clear();
  adj[sink].clear();
  int tt = 0, maxallow = k;
  for (int i = 0; i < n; i++) {
    adj[i].clear();
    res[i][sink] = res[sink][i] = res[src][i] = res[i][src] = 0;
    if (pt[i] == pt[v] || pt[i] == pt[u]) {
      involve[i] = 2;
      cnt++;
      continue;
    }
    if (dist2(i, v) > d2 || dist2(i, u) > d2) {
      involve[i] = 0;
      maxallow--;
      continue;
    }
    involve[i] = 1;
    if (cross(pt[v], pt[u], pt[i]) < 0)
      append(src, i);
    else
      append(i, sink);
    tt++;
  }
  fid++;
  for (int i = 0; i < n; i++) {
    if (involve[i] != 1) continue;
    for (int j = i + 1; j < n; j++) {
      if (involve[j] != 1) continue;
      if (dist2(i, j) <= d2) continue;
      if (res[src][i] + res[i][src])
        append(i, j);
      else
        append(j, i);
    }
  }
  for (int i = 0; i < n; i++) {
    if (adj[i].size() >= k + 1) {
      forbidden[i] = fid;
      tt--;
      maxallow--;
    }
  }
  if (maxallow < 0) return -1;
  if (tt + cnt < n - k) return -1;
  int ret = tt - flow(max(0, maxallow)) + cnt;
  return ret;
}
inline void solve() {
  vector<Edge> e;
  for (int v = 0; v < n; v++)
    for (int u = v + 1; u < n; u++) e.push_back(Edge(v, u, dist2(v, u)));
  sort(e.begin(), e.end());
  int i;
  int maxtry = min((int)e.size(), k * (k + 1) + 1), lasti = -1;
  for (int i = (int)e.size() - 1; i >= 0; i--) {
    if (failcnt[e[i].v] > k) continue;
    if (failcnt[e[i].u] > k) continue;
    if (maxtry-- <= 0) break;
    int ret = test(e[i].v, e[i].u);
    if (ret >= n - k) lasti = i;
    failcnt[e[i].v]++;
    failcnt[e[i].u]++;
  }
  int v = e[lasti].v, u = e[lasti].u;
  int ret = test(e[lasti].v, e[lasti].u);
  int rem = k;
  for (int j = 0; j < n; j++) {
    if (n - j <= rem) {
      printf("%d ", j + 1);
      rem--;
      continue;
    }
    if (involve[j] == 2) continue;
    if (involve[j] == 0 || forbidden[j] == fid || dist2(j, v) > dist2(v, u) ||
        dist2(j, u) > dist2(v, u) ||
        (res[src][j] + res[j][src]) && visited[j] < visid ||
        (res[sink][j] + res[j][sink]) && visited[j] == visid) {
      printf("%d ", j + 1);
      rem--;
    }
  }
  puts("");
}
int main(void) {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) pt[i].input();
  solve();
  return 0;
}
