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"
#define YES "YES"
#define NO "NO"
#define YESNO OUT(three(solve(),YES,NO))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(b);i>=(LL)(a);i--)
#define REP(i,b) FOR(i,zero,b)
#define rep REP
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,b,zero)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB push_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define zero 0LL
#define all ALL
#define pb emplace_back
#define eb pb
#define int long long
using namespace std;
template<typename T> inline void maximize(T &a, T b) { a = max(a, b); }
template<typename T> inline void minimize(T &a, T b) { a = min(a, b); }
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef int LL;
typedef double ld;
typedef int ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef tuple<int, int, int, int> tapu;
typedef vector<tapu> Vtapu;
typedef priority_queue<tapu, Vtapu, greater<tapu> > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
const int SIZE1 = 3e5 + 1000;
const int SIZE2 = 5010;
const int SIZE3 = 430;
const int SIZE = SIZE1;
const int MAPSIZE = 40;
const LL p = 7 + 1e9;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
typedef pair<ld, ut> pld;
 
LL N;
LL X[SIZE],Y[SIZE];
ld solve(VI v){
  ld dir=0;
  ld dirsum=0;
  v.pb(0);  
  pr now=pr(0,0);
  REP(i,N+1){
    ld ndir=atan2(Y[v[i]]-now.second,X[v[i]]-now.first);
        
    dirsum+=min(2*M_PI-abs<ld>(ndir-dir),abs<ld>(ndir-dir));
  //  cout << dir*360/(2*M_PI) << endl;
    dir=ndir;  
    now=pr(X[v[i]],Y[v[i]]);
  }
  //cout << dirsum << endl;
  //outputVI(v);
  return dirsum;
}
signed main(){
  cout << setprecision(10) << endl;
  cin >> N;
  VI v;
  EFOR(i,1,N){
    cin >> X[i] >> Y[i];
  }
  EFOR(i,1,N) v.pb(i);
  ld ans=INF;
  do{
    MN(ans,solve(v));
  }while(next_permutation(ALL(v)));
  //cout << ans << endl;
  cout << ans*360*1.0/(2*M_PI) << endl;
  return 0;
}