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>
void Get(int &T) {
  char C;
  bool F = 0;
  for (; C = getchar(), C < '0' || C > '9';)
    if (C == '-') F = 1;
  for (T = C - '0'; C = getchar(), C >= '0' && C <= '9'; T = T * 10 + C - '0')
    ;
  F && (T = -T);
}
int N, K;
int X[1005];
int Y[1005];
int Dis[1005][1005];
int Distance[1000005];
int DistanceTop;
void Init() {
  Get(N);
  Get(K);
  for (int i = 1; i <= N; i++) {
    Get(X[i]);
    Get(Y[i]);
  }
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++) {
      int DX = X[i] - X[j];
      int DY = Y[i] - Y[j];
      Dis[i][j] = DX * DX + DY * DY;
      Distance[++DistanceTop] = Dis[i][j];
    }
  std::sort(Distance + 1, Distance + DistanceTop + 1);
  int j = 1;
  for (int i = 2; i <= DistanceTop; i++) {
    if (Distance[i] != Distance[i - 1]) Distance[++j] = Distance[i];
  }
  DistanceTop = j;
}
int Delete[1005];
struct E {
  int V, Next;
} Edge[1000005];
int EdgeTop;
int Head[1005];
void Add(int U, int V) {
  Edge[++EdgeTop] = (E){V, Head[U]};
  Head[U] = EdgeTop;
}
bool DFS(int X, int Last) {
  if (X > N) return 1;
  if (Delete[X]) return DFS(X + 1, Last);
  int NewLast = Last;
  for (int i = Head[X]; i; i = Edge[i].Next) {
    if (Delete[Edge[i].V] == 0) NewLast--;
    Delete[Edge[i].V]++;
  }
  if (NewLast >= 0 && DFS(X + 1, NewLast) == 1) return 1;
  for (int i = Head[X]; i; i = Edge[i].Next) {
    Delete[Edge[i].V]--;
  }
  Last--;
  if (NewLast < Last) {
    Delete[X]++;
    if (Last >= 0 && DFS(X + 1, Last) == 1) return 1;
    Delete[X]--;
  }
  return 0;
}
bool Check(int X) {
  EdgeTop = 0;
  for (int i = 1; i <= N; i++) Delete[i] = Head[i] = 0;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      if (Dis[i][j] > X) {
        Add(i, j);
      }
    }
  }
  return DFS(1, K);
}
int Bin(int L, int R) {
  for (;;) {
    if (L == R) return L;
    if (L + 1 == R) {
      if (Check(Distance[L])) return L;
      return R;
    }
    int Mid = L + R;
    Mid >>= 1;
    if (Check(Distance[Mid])) {
      R = Mid;
    } else
      L = Mid;
  }
}
int Ans;
void Work() {
  Ans = Bin(1, DistanceTop);
  Ans = Distance[Ans];
  Check(Ans);
}
void Output() {
  int k = 1;
  for (int i = 1; i <= N; i++)
    if (Delete[i] > 0) {
      printf("%d ", i);
      k++;
    }
  for (int i = 1; i <= N && k <= K; i++)
    if (Delete[i] == 0) {
      printf("%d ", i);
      k++;
    }
}
int main() {
  Init();
  Work();
  Output();
  return 0;
}