#include <bits/stdc++.h>
using namespace std;
int n,i,j,k,a[5050],x[5050],y[5050];
long long scp(int i, int j, int k) {
  long long xa=x[i]-x[j];
  long long ya=y[i]-y[j];
  long long xb=x[k]-x[j];
  long long yb=y[k]-y[j];
  return xa*xb+ya*yb;
}
int main() {
  scanf("%d",&n);
  for (i=0; i<n; i++) scanf("%d%d",&x[i],&y[i]);
  a[0]=0;
  a[1]=1;
  for (i=2; i<n; i++) {
    for (j=i; j>=0; --j) {
      if (j>=2 && scp(a[j-2],a[j-1],i)<=0) continue;
      if (j>=1 && j<i && scp(a[j-1],i,a[j])<=0) continue;
      if (j+1<i && scp(i,a[j],a[j+1])<=0) continue;
      for (k=i; k>j; --k) a[k]=a[k-1];
      a[j]=i;
      break;
    }
    if (j<0) { puts("-1"); return 0; }
  }
  for (i=0; i<n; i++) printf("%d ",a[i]+1);
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

Nezzar loves the game osu!.

osu! is played on beatmaps, which can be seen as an array consisting of distinct points on a plane. A beatmap is called nice if for any three consecutive points A,B,C listed in order, the angle between these three points, centered at B, is strictly less than 90 degrees.

<image> Points A,B,C on the left have angle less than 90 degrees, so they can be three consecutive points of a nice beatmap; Points A',B',C' on the right have angle greater or equal to 90 degrees, so they cannot be three consecutive points of a nice beatmap.

Now Nezzar has a beatmap of n distinct points A_1,A_2,…,A_n. Nezzar would like to reorder these n points so that the resulting beatmap is nice.

Formally, you are required to find a permutation p_1,p_2,…,p_n of integers from 1 to n, such that beatmap A_{p_1},A_{p_2},…,A_{p_n} is nice. If it is impossible, you should determine it.

Input

The first line contains a single integer n (3 ≤ n ≤ 5000).

Then n lines follow, i-th of them contains two integers x_i, y_i (-10^9 ≤ x_i, y_i ≤ 10^9) — coordinates of point A_i.

It is guaranteed that all points are distinct.

Output

If there is no solution, print -1.

Otherwise, print n integers, representing a valid permutation p.

If there are multiple possible answers, you can print any.

Example

Input


5
0 0
5 0
4 2
2 1
3 0


Output


1 2 5 3 4

Note

Here is the illustration for the first test:

<image>

Please note that the angle between A_1, A_2 and A_5, centered at A_2, is treated as 0 degrees. However, angle between A_1, A_5 and A_2, centered at A_5, is treated as 180 degrees.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n,i,j,k,a[5050],x[5050],y[5050];
long long scp(int i, int j, int k) {
  long long xa=x[i]-x[j];
  long long ya=y[i]-y[j];
  long long xb=x[k]-x[j];
  long long yb=y[k]-y[j];
  return xa*xb+ya*yb;
}
int main() {
  scanf("%d",&n);
  for (i=0; i<n; i++) scanf("%d%d",&x[i],&y[i]);
  a[0]=0;
  a[1]=1;
  for (i=2; i<n; i++) {
    for (j=i; j>=0; --j) {
      if (j>=2 && scp(a[j-2],a[j-1],i)<=0) continue;
      if (j>=1 && j<i && scp(a[j-1],i,a[j])<=0) continue;
      if (j+1<i && scp(i,a[j],a[j+1])<=0) continue;
      for (k=i; k>j; --k) a[k]=a[k-1];
      a[j]=i;
      break;
    }
    if (j<0) { puts("-1"); return 0; }
  }
  for (i=0; i<n; i++) printf("%d ",a[i]+1);
  return 0;
}

```