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.

Examples

Input

3
7 12 8


Output

4


Input

2
11 11


Output

2


Input

1
0


Output

0
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<functional>

using namespace std;

int pow2(int a) {
	if (a == 0) { return 1; }
	return 2 * pow2(a - 1);
}

int main() {
	int N, D, G[13];
	int cn, P[13];
	int k,M[13],L[13];
	int s, ans, last;

	scanf("%d", &N);
	
	for (int i = 0; i < 13; i++) {
		G[i] = 0;
	}


	for (int i = 0; i < N; i++) {
		scanf("%d", &D);
		G[D]++;
	}

	if (G[0] > 0) { printf("0\n"); return 0; }
	for (int i = 1; i < 12; i++) {
		if (G[i] > 2) {
			printf("0\n");
			return 0;
		}
	}
	if (G[12] > 1) { printf("0\n"); return 0;}

	cn = 0;
	for (int i = 1; i < 12; i++) {
		if (G[i] == 1) {
			P[cn] = i;
			cn++;
		}
	}

	ans = 0;
	for (int i = 0; i < pow2(cn); i++) {
		k = i;
		for (int j = 0; j < cn; j++) {
			M[j] = k % 2;/*M[j]が0なら午前*/
			k = k / 2;
		}

		for (int j = 0; j < 24; j++) {
			L[j] = 0;
		}

		L[0] = 1;
		for (int j = 1; j < 12; j++) {
			if (G[j] == 2) {
				L[j] = 1; L[24 - j] = 1;
			}
		}
		if (G[12] == 1) { L[12] = 1; }
	
		for (int j = 0; j < cn; j++) {
	
			if (M[j] == 0) { L[P[j]] = 1; }
			else { L[24 - P[j]] = 1; }
		}


		s = 24;
		last = 0;
		for (int j = 1; j < 24; j++) {
			if (L[j] == 1) { 
				if (s > j - last) { s = j - last; } 
				last = j;
			}
		}
		if (s > 24 - last) { s = 24 - last; }
		if (ans < s) { ans = s; }
		
	}

	printf("%d\n", ans);
	
}