#include <bits/stdc++.h>

using namespace std;

long long ccw(pair<long long, long long> x, pair<long long, long long> y, pair<long long, long long> z){
	y.first -= x.first, y.second -= x.second;
	z.first -= x.first, z.second -= x.second;
	long long val = y.first * z.second - y.second * z.first;
	if(val > 0) return 1;
	if(val < 0) return -1;
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	long long N;
	cin >> N;
	vector<pair<long long, long long>> a(N);
	for(long long i = 0; i < N; ++i){
		cin >> a[i].first >> a[i].second;
	}
	vector<long long> chk(N), ans(N);
	ans[0] = 0;
	chk[0] = 1;
	for(long long i = 1; i < N; ++i){
		long long pos = ans[i - 1];
		long long num = -1;
		for(long long j = 0; j < N; ++j){
			if(chk[j]){
				continue;
			}
			if(num == -1){
				num = j;
				continue;
			}
			if(ccw(a[num], a[j], a[pos]) > 0){
				pair<long long, long long> p = {a[num].first - a[num].second + a[pos].second, a[num].second + a[num].first - a[pos].first};
				if(ccw(a[num], a[j], p) >= 0){
					num = j;
				}
			}
			else if(ccw(a[num], a[j], a[pos]) < 0){
				pair<long long, long long> p = {a[num].first + a[num].second - a[pos].second, a[num].second - a[num].first + a[pos].first};
				if(ccw(a[num], a[j], p) <= 0){
					num = j;
				}
			}
			else{
				if(abs(a[num].first - a[pos].first) < abs(a[j].first - a[pos].first) || 
					abs(a[num].second - a[pos].second) < abs(a[j].second - a[pos].second)){
					num = j;
				}
			}
		}
		chk[num] = 1;
		ans[i] = num;
	}
	for(long long i = 0; i < N; ++i){
		cout << ans[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;

long long ccw(pair<long long, long long> x, pair<long long, long long> y, pair<long long, long long> z){
	y.first -= x.first, y.second -= x.second;
	z.first -= x.first, z.second -= x.second;
	long long val = y.first * z.second - y.second * z.first;
	if(val > 0) return 1;
	if(val < 0) return -1;
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	long long N;
	cin >> N;
	vector<pair<long long, long long>> a(N);
	for(long long i = 0; i < N; ++i){
		cin >> a[i].first >> a[i].second;
	}
	vector<long long> chk(N), ans(N);
	ans[0] = 0;
	chk[0] = 1;
	for(long long i = 1; i < N; ++i){
		long long pos = ans[i - 1];
		long long num = -1;
		for(long long j = 0; j < N; ++j){
			if(chk[j]){
				continue;
			}
			if(num == -1){
				num = j;
				continue;
			}
			if(ccw(a[num], a[j], a[pos]) > 0){
				pair<long long, long long> p = {a[num].first - a[num].second + a[pos].second, a[num].second + a[num].first - a[pos].first};
				if(ccw(a[num], a[j], p) >= 0){
					num = j;
				}
			}
			else if(ccw(a[num], a[j], a[pos]) < 0){
				pair<long long, long long> p = {a[num].first + a[num].second - a[pos].second, a[num].second - a[num].first + a[pos].first};
				if(ccw(a[num], a[j], p) <= 0){
					num = j;
				}
			}
			else{
				if(abs(a[num].first - a[pos].first) < abs(a[j].first - a[pos].first) || 
					abs(a[num].second - a[pos].second) < abs(a[j].second - a[pos].second)){
					num = j;
				}
			}
		}
		chk[num] = 1;
		ans[i] = num;
	}
	for(long long i = 0; i < N; ++i){
		cout << ans[i] + 1 << ' ';
	}
	return 0;
}
```