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: Accepted)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

#define dprint(Exp,...) if(Exp){fprintf(stderr, __VA_ARGS__);}
#define printe(...) fprintf(stderr, __VA_ARGS__);
#define PrtExp(_Exp)  cerr<< #_Exp <<" = "<< (_Exp)
#define PrtExpN(_Exp)  cerr<< #_Exp <<" = "<< (_Exp) <<"\n"

#define SINT(n) scanf("%d",&n)
#define SINT2(n,m) scanf("%d %d",&n,&m)
#define SINT3(n,m,o) scanf("%d %d %d",&n,&m,&o)
#define SINT4(n,m,o,p) scanf("%d %d %d %d",&n,&m,&o,&p)
#define SINT5(n,m,o,p,q) scanf("%d %d %d %d %d",&n,&m,&o,&p,&q)
#define SLL(n) scanf("%lld",&n)
#define SLL2(n,m) scanf("%lld %lld",&n,&m)
#define SLL3(n,m,o) scanf("%lld %lld %lld",&n,&m,&o)
#define SST(s) scanf("%s",s)
#define SCH(c) scanf("%c",&c)

#define GC() getchar()

#define PINT(n) printf("%d",(int)(n))
#define PINT2(n,m) printf("%d %d",(int)(n),(int)(m))
#define PINT3(n,m,l) printf("%d %d %d",(int)(n),(int)(m),(int)(l))
#define PLL(n) printf("%lld",(long long)(n))
#define PST(s) printf("%s",(s))
#define PCH(s) printf("%c",(s))

#define PINTN(n) printf("%d\n",(int)(n))
#define PINT2N(n,m) printf("%d %d\n",(int)(n),(int)(m))
#define PINT3N(n,m,l) printf("%d %d %d\n",(int)(n),(int)(m),(int)(l))
#define PLLN(n) printf("%lld\n",(long long)(n))
#define PSTN(s) printf("%s\n",(s))
#define PCHN(s) printf("%c\n",(s))

#define PSP() printf(" ")
#define PN() printf("\n")

#define PC(c) putchar(c)
#define CSP (' ')
#define SN ("\n")

#define rep(i,a) for(int i=0;i<a;i++)
#define reP(i,a) for(int i=0;i<=a;i++)
#define Rep(i,a) for(int i=a-1;i>=0;i--)
#define ReP(i,a) for(int i=a;i>=0;i--)

#define rEp(i,a) for(i=0;i<a;i++)
#define rEP(i,a) for(i=0;i<=a;i++)
#define REp(i,a) for(i=a-1;i>=0;i--)
#define REP(i,a) for(i=a;i>=0;i--)

#define repft(i,a,b) for(int i=a;i<b;i++)
#define repfT(i,a,b) for(int i=a;i<=b;i++)
#define Repft(i,a,b) for(int i=a-1;i>=b;i--)
#define RepfT(i,a,b) for(int i=a;i>=b;i--)

#define foreach(a,it) for(auto it = a.begin(); it != a.end(); ++it)

#define FILL(a,v) fill(begin(a),end(a), v)
#define FILL0(a) memset(a,0,sizeof(a))
#define FILL1(a) memset(a,-1,sizeof(a))

typedef long long ll;

typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll>   Pll;


const int INF = 0x2aaaaaaa; // 715,827,882
const ll INFLL = 0x1f1f1f1f1f1f1f1fLL;//2,242,545,357,980,376,863

template <class A, class B> inline ostream& operator<<(ostream& st, const pair<A, B>& P) { return st << "(" << P.first << "," << P.second << ")"; };
template <class A, class B> inline pair<A, B> operator+(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first + Q.first, P.second + Q.second); };
template <class A, class B> inline pair<A, B> operator-(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first - Q.first, P.second - Q.second); };

#define fs  first
#define sc  second


int a[13] = {};
int b[50];

int main() {
	ios::sync_with_stdio(0);

	int n;
	cin >> n;
	rep(i, n) {
		int d;
		cin >> d;
		a[d]++;
	}
	a[0]++;

	vector<int> v, w;
	rep(i, 13) {
		if (a[i] == 1)v.push_back(i);
		if (a[i] == 2)w.push_back(i);
		if (a[i] > 2) {
			cout << 0 << endl;
			return 0;
		}
	}
	if (a[0] > 1) {
		cout << 0 << endl;
		return 0;
	}

	int si = v.size();
	int wsi = w.size();

	int ret = 0;

	rep(pat, 1<<si){
		rep(i, si) {
			if (pat& (1 << i)) b[i] = v[i];
			else               b[i] = 24 - v[i];
		}
		rep(i, wsi) {
			b[si + i * 2]   = w[i];
			b[si + i * 2+1] = 24-w[i];
		}
		int r = 24;
		rep(i, si+wsi*2) {
			repft(j, i + 1, si+wsi*2) {
				int d = b[i] - b[j];
				if (d < 0) d = -d;
				if (d > 12) d = 24 - d;
				r = min(r, d);
			}
		}
		ret = max(r, ret);
	}

	cout << ret << endl;


	return 0;
}