void dfs(int x, int y, int c) {
	if(x || y) { //不为起点
		if(!c) dfs(x - 1, y, pre[x][y][c]); //现在在左边，递归到上一个位置
		else dfs(x, y - 1, pre[x][y][c]); //现在在右边
	}
	printf("%d ", (c) ? po(y) + 1 : pr(x) + 1); //要倒着输出，所以放在后面
}

#include<bits/stdc++.h>
using namespace std;
int n;
struct node {
  double x, y;
  int bian;
}a[1010]; //记录点地坐标
double f[1010][1010][2];
int pre[1010][1010][2];
int st;
inline double dis(int x, int y) { //两个点的距离
  return sqrt((a[x].x - a[y].x) * (a[x].x - a[y].x) + (a[x].y - a[y].y) * (a[x].y - a[y].y));
}
//分别是后第x的序号和前第x的序号
#define po(x) ((st + n - (x)) % n) //这里x千万要打括号！血与泪的教训！万恶的宏！
#define pr(x) ((st + x) % n)
int minn, minnc;
void dfs(int x, int y, int c) { //递归输出答案
	if(x || y) {
		if(!c) dfs(x - 1, y, pre[x][y][c]);
		else dfs(x, y - 1, pre[x][y][c]);
	}
	printf("%d ", (c) ? po(y) + 1 : pr(x) + 1); //由于我是从0~n-1所以要+1
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lf %lf", &a[i].x, &a[i].y);
    a[i].bian = i;
    if (a[i].y > a[st].y) st = i;
  }
  for (int i = 0; i < n; ++i) {
  	for (int j = 0; j < n; ++j) {
  		f[i][j][0] = f[i][j][1] = 1e18; //都初始化为无穷大
		}
	}
  f[0][0][0] = f[0][0][1] = 0.0; //起点为0
  //先算出有一个为0的情况（也可以不写在下面特判，但是注意数组越界）
  for (int i = 1; i < n; ++i) {
    f[i][0][0] = f[i - 1][0][0] + dis(pr(i - 1), pr(i));
    pre[i][0][0] = 0;
  }
  for (int i = 1; i < n; ++i) {
    f[0][i][1] = f[0][i - 1][1] + dis(po(i - 1), po(i));
    pre[0][i][1] = 1;
  }
  for (int i = 1; i < n; ++i) {
    for (int j = 1; j < n - i; ++j) {
    	double tmp1 = f[i - 1][j][0] + dis(pr(i - 1), pr(i)), tmp2 = f[i - 1][j][1] + dis(po(j), pr(i)); //暂时存一下两种情况
      f[i][j][0] = min(tmp1, tmp2); //取最小值
      if (f[i][j][0] == tmp1) pre[i][j][0] = 0; //判断从哪边过来
      else pre[i][j][0] = 1;
      tmp1 = f[i][j - 1][1] + dis(po(j - 1), po(j)), tmp2 = f[i][j - 1][0] + dis(pr(i), po(j)); //同理
      f[i][j][1] = min(tmp1, tmp2);
      if (f[i][j][1] == tmp1) pre[i][j][1] = 1;
      else pre[i][j][1] = 0;
    }
  }
  //找最小的
  for (int i = 0; i < n; ++i) {
  	if (f[i][n - i - 1][0] < f[minn][n - minn - 1][minnc]) minn = i, minnc = 0;
  	if (f[i][n - i - 1][1] < f[minn][n - minn - 1][minnc]) minn = i, minnc = 1;
  }
  dfs(minn, n - minn - 1, minnc); //递归输出答案
  return 0;
}