#include <bits/stdc++.h>
using namespace std;
const int maxn = 1100;
int n, K, m, D[maxn * maxn];
struct Tpnt {
  int x, y;
} A[maxn];
inline int sqr(int x) { return x * x; }
inline int dist(const Tpnt &a, const Tpnt &b) {
  return sqr(a.x - b.x) + sqr(a.y - b.y);
}
namespace Ninit {
void init() {
  int i, j;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; ++i) scanf("%d%d", &A[i].x, &A[i].y);
  for (i = 1; i < n; ++i)
    for (j = i + 1; j <= n; ++j) D[m++] = dist(A[i], A[j]);
  sort(D, D + m), m = unique(D, D + m) - D;
}
}  // namespace Ninit
namespace Nsolve {
namespace Ncheck {
int U[maxn];
vector<int> E[maxn];
bool dfs(int x, int R) {
  int i, k;
  vector<int>::iterator p;
  if (x > n) return true;
  if (E[x].empty() || U[x] <= n) return dfs(x + 1, R);
  {
    for (k = 0, p = E[x].begin(); p != E[x].end(); ++p)
      k += (U[*p] == n + 1 ? (U[*p] = x, 1) : 0);
    if (R >= k && dfs(x + 1, R - k)) return true;
    for (p = E[x].begin(); p != E[x].end(); ++p) U[*p] == x ? U[*p] = n + 1 : 0;
    if (k < 2) return false;
  }
  {
    U[x] = x;
    if (R && dfs(x + 1, R - 1)) return true;
    U[x] = n + 1;
  }
  return false;
}
bool Chk(int Lmt) {
  int i, j;
  for (i = 1; i <= n; ++i) E[i].clear(), U[i] = n + 1;
  for (i = 1; i < n; ++i)
    for (j = i + 1; j <= n; ++j)
      if (dist(A[i], A[j]) > Lmt) E[i].push_back(j);
  return dfs(1, K);
}
}  // namespace Ncheck
using Ncheck::Chk;
void solve() {
  int i, l, r, mid;
  for (l = -1, r = m - 1; l + 1 < r; Chk(D[mid]) ? r = mid : l = mid)
    mid = l + r >> 1;
  for (Chk(D[r]), i = 1; i <= n; ++i)
    if (Ncheck::U[i] <= n) printf("%d ", i), --K;
  for (i = 1; K && i <= n; ++i)
    if (Ncheck::U[i] > n) printf("%d ", i), --K;
  printf("\n");
}
}  // namespace Nsolve
int main() {
  Ninit::init();
  Nsolve::solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1100;
int n, K, m, D[maxn * maxn];
struct Tpnt {
  int x, y;
} A[maxn];
inline int sqr(int x) { return x * x; }
inline int dist(const Tpnt &a, const Tpnt &b) {
  return sqr(a.x - b.x) + sqr(a.y - b.y);
}
namespace Ninit {
void init() {
  int i, j;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; ++i) scanf("%d%d", &A[i].x, &A[i].y);
  for (i = 1; i < n; ++i)
    for (j = i + 1; j <= n; ++j) D[m++] = dist(A[i], A[j]);
  sort(D, D + m), m = unique(D, D + m) - D;
}
}  // namespace Ninit
namespace Nsolve {
namespace Ncheck {
int U[maxn];
vector<int> E[maxn];
bool dfs(int x, int R) {
  int i, k;
  vector<int>::iterator p;
  if (x > n) return true;
  if (E[x].empty() || U[x] <= n) return dfs(x + 1, R);
  {
    for (k = 0, p = E[x].begin(); p != E[x].end(); ++p)
      k += (U[*p] == n + 1 ? (U[*p] = x, 1) : 0);
    if (R >= k && dfs(x + 1, R - k)) return true;
    for (p = E[x].begin(); p != E[x].end(); ++p) U[*p] == x ? U[*p] = n + 1 : 0;
    if (k < 2) return false;
  }
  {
    U[x] = x;
    if (R && dfs(x + 1, R - 1)) return true;
    U[x] = n + 1;
  }
  return false;
}
bool Chk(int Lmt) {
  int i, j;
  for (i = 1; i <= n; ++i) E[i].clear(), U[i] = n + 1;
  for (i = 1; i < n; ++i)
    for (j = i + 1; j <= n; ++j)
      if (dist(A[i], A[j]) > Lmt) E[i].push_back(j);
  return dfs(1, K);
}
}  // namespace Ncheck
using Ncheck::Chk;
void solve() {
  int i, l, r, mid;
  for (l = -1, r = m - 1; l + 1 < r; Chk(D[mid]) ? r = mid : l = mid)
    mid = l + r >> 1;
  for (Chk(D[r]), i = 1; i <= n; ++i)
    if (Ncheck::U[i] <= n) printf("%d ", i), --K;
  for (i = 1; K && i <= n; ++i)
    if (Ncheck::U[i] > n) printf("%d ", i), --K;
  printf("\n");
}
}  // namespace Nsolve
int main() {
  Ninit::init();
  Nsolve::solve();
  return 0;
}

```