Problem:
Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
typedef  unsigned long long int ull;
typedef pair<ll,ll> P;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
//const ll mod=998244353;
const ll mod=1000000007;
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
ll gcd(ll a,ll b) {return b ? gcd(b,a%b):a;}
ll lcm(ll c,ll d){return c/gcd(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int main(){
int n;cin>>n;
V<int> a(n);
for(int i=0;i<n;i++)cin>>a[i];
if(n>24){
    cout<<0<<endl;
    return 0;
}
if(n>12){
    V<int> cnt(n);
    cnt[0]++;
    for(int i=0;i<n;i++){
        cnt[a[i]]++;
        if(cnt[a[i]]==3){
            cout<<0<<endl;
            return 0;
        }
    }
    cout<<1<<endl;
    return 0;
}
int ans=0;
for(int bit=0;bit<(1<<n);bit++){
    V<int> d(n+1);
    for(int i=0;i<n;i++){
        if(a[i]==0||bit&(1<<i))d[i]=a[i];
        else d[i]=24-a[i];
    }
    d[n]=0;
    int res=INT_MAX;
    for(int i=0;i<n+1;i++){
        for(int j=i+1;j<n+1;j++){
              chmin(res,abs(d[i]-d[j]));
        }
    }
    chmax(ans,res);
}
cout<<ans<<endl;
}