Problem: Problem

Satake doesn't like being bent.
For example, I don't like pencils, chopsticks, and bent-shaped houses, and I don't like taking actions such as turning right or left.

By the way, Satake who lives on the XY plane is $ N $ shops $ (X_ {1}, Y_ {1}), (X_ {2}, Y_ {2}), \ ldots, (X_ {N} , Y_ {N}) I was asked to shop for $. Depart from the house at coordinates $ (0, 0) $ in the direction of $ (1, 0) $, shop at all stores, and then return home. You can go around the shops in any order. As a user, Satake can perform any of the following actions as many times as you like.

1. Go as much as you like in the direction you are facing
2. When the coordinates are the same as the shop or house, rotate clockwise or counterclockwise by the desired angle on the spot.
3. When the coordinates are the same as the store, shop while maintaining the coordinates and the direction in which they are facing.



As I said earlier, Satake doesn't like to bend. To prepare your mind, find the minimum sum of the angles that Satake will rotate in Action 2. As an example, if you rotate $ 90 ^ {\ circ} $ clockwise and then $ 90 ^ {\ circ} $ counterclockwise, the sum of the angles will be $ 180 ^ {\ circ} $.

Constraints

The input meets the following conditions:

* $ 2 \ le N \ le 8 $
* $ -1000 \ le X_ {i}, Y_ {i} \ le 1000 \ quad (1 \ le i \ le N) $
* $ (X_ {i}, Y_ {i}) \ ne (0, 0) \ quad (1 \ le i \ le N) $
* $ (X_ {i}, Y_ {i}) \ ne (X_ {j}, Y_ {j}) \ quad (i \ ne j) $
* All inputs are integers

Input

The input is given in the following format:


$ N $
$ X_ {1} $ $ Y_ {1} $
$ \ vdots $
$ X_ {N} $ $ Y_ {N} $


The input consists of $ N + 1 $ lines.
The $ 1 $ line is given $ N $, which represents the number of stores to shop.
The $ N $ line starting from the $ 2 $ line is given the coordinates $ X_ {i}, Y_ {i} $ of the store where you shop, separated by blanks.

Output

Please output the minimum value of the sum of the angles that Satake rotates in degrees. However, the correct answer is given only when the absolute error from the assumed solution is $ 10 ^ {-4} $ or less.

Examples

Input

2
0 1
0 -1


Output

450.00000000


Input

3
1 0
0 1
-2 -1


Output

386.565051

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
const int INF=1e9,MOD=1e9+7,ohara=1e6+10;
const ll LINF=1e18;
using namespace std;
    
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define Size(n) (n).size()
#define Cout(x) cout<<(x)<<endl
#define doublecout(a) cout<<fixed<<setprecision(15)<<a<<endl;
#define Cerr(x) cerr<<(x)<<endl
#define fi first
#define se second
#define P pair<ll,ll> 
#define m_p make_pair
#define V vector<ll> 
#define U_MAP unordered_map<ll,ll>
#define pq priority_queue<ll>
#define rpq priority_queue<ll,vector<ll>,greater<ll>>
#define p_b push_back
    
ll n,cnt,a,b,c,d,tmpp,m,h,w,x[20],y[20],sum,pos,k;
ld doua;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
//int dy[]={-1,0,1,-1,1,-1,0,1};
//int dx[]={-1,-1,-1,0,0,1,1,1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool fl;
struct edge{int to,cost;};
double ans=-1.0;
    
//-------------------------↓↓↓↓↓↓------------------------
    
//------ 自分を天才だと信じる ------
    
int main(void){
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    
    cin>>n;
    rep(i,n)cin>>x[i]>>y[i];
    vector<ll> p(n);
    rep(i,n)p[i]=i;
    do{
        double theta=0.0;
        double anss=0.0;
        ll bex=0,bey=0;
        rep(i,n){
            double nowtheta;
            if(x[p[i]]==bex){
                //Cout("A");
                if(y[p[i]]>bey){
                    nowtheta=90.0;
                }
                else{
                    nowtheta=270.0;
                }
            }
            else if(y[p[i]]==bey){
                if(x[p[i]]>bex){
                    nowtheta=0.0;
                }
                else{
                    nowtheta=180.0;
                }
            }
            else{
                //Cout("A");
                double tmp;
                tmp=(double)abs(y[p[i]]-bey)/(double)abs(x[p[i]]-bex);
                nowtheta=atan(tmp);
                nowtheta=nowtheta*180.0/M_PI;
                //doublecout(nowtheta/M_PI*180.0);
                if(x[p[i]]<bex&&y[p[i]]>bey){
                    nowtheta=180.0-nowtheta;
                }
                else if(x[p[i]]<bex&&y[p[i]]<bey){
                    nowtheta=nowtheta+180.0;
                }
                else if(x[p[i]]>bex&&y[p[i]]<bey){
                    nowtheta=360.0-nowtheta;
                }
            }
            anss+=min(abs(theta-nowtheta),360.0-abs(theta-nowtheta));
            theta=nowtheta;
            bex=x[p[i]];
            bey=y[p[i]];
            //Cout(anss);
            //cout<<bex<<" "<<bey<<"\n";
        }
        //cout<<"\n";
        double nowtheta;
            if(0==bex){
                //Cout("A");
                if(0>bey){
                    nowtheta=90.0;
                }
                else{
                    nowtheta=270.0;
                }
            }
            else if(0==bey){
                if(0>bex){
                    nowtheta=0.0;
                }
                else{
                    nowtheta=180.0;
                }
            }
            else{
                //Cout("A");
                double tmp;
                tmp=(double)abs(0-bey)/(double)abs(0-bex);
                nowtheta=atan(tmp);
                nowtheta=nowtheta*180.0/M_PI;
                if(0<bex&&0>bey){
                    nowtheta=180.0-nowtheta;
                }
                else if(0<bex&&0<bey){
                    nowtheta=nowtheta+180.0;
                }
                else if(0>bex&&0<bey){
                    nowtheta=360.0-nowtheta;
                }
            }
           // Cout(anss);
            anss+=min(abs(theta-nowtheta),360.0-abs(theta-nowtheta));
            theta=nowtheta;
        //cout<<x[p[1]]<<" "<<y[p[1]]<<"\n";
        //Cout(anss);
        //cout<<"\n";
        if(ans==-1.0)ans=anss;
        else ans=min(ans,anss);
    }while(next_permutation(all(p)));
    doublecout(ans);
    return 0;
}