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 "bits/stdc++.h"
using namespace std;

int main() {
	int N, D, X[13] = {}, ANS = 24;
	vector<int> V;
	bool L = true, R = true, Y[24] = {};
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> D;
		X[D]++;
	}
	if (X[0] != 0) {
		cout << 0;
		return 0;
	}
	for (int i = 1; i < 12; i++) {
		if (X[i] > 2) {
			cout << 0;
			return 0;
		}
		if (X[i] == 2) {
			Y[i] = true, Y[24 - i] = true;
			L = true, R = true;
		}
		if (X[i] == 1) {
			if (L) {
				Y[i] = true;
				L = false, R = true;
			}
			else {
				Y[24 - i] = true;
				L = true, R = false;
			}
		}
	}
	if (X[12] > 1) {
		cout << 0;
		return 0;
	}
	if (X[12] == 1) Y[12] = true;
	V.push_back(0);
	for (int i = 0; i < 24; i++) {
		if (Y[i]) V.push_back(i);
	}
	V.push_back(V[0] + 24);
	for (int i = 0; i < V.size() - 1; i++) {
		ANS = min(ANS, V[i + 1] - V[i]);
	}
	cout << ANS;
}