Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.

Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.

You know that there will be n interesting minutes t1, t2, ..., tn. Your task is to calculate for how many minutes Limak will watch the game.

The first line of the input contains one integer n (1 ≤ n ≤ 90) — the number of interesting minutes.

The second line contains n integers t1, t2, ..., tn (1 ≤ t1 < t2 < ... tn ≤ 90), given in the increasing order.

Print the number of minutes Limak will watch the game.

In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.

In the second sample, the first 15 minutes are boring.

In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game.

题意：

给这么多时间点,这些时间点是interesting的点,如果连续15分钟不出现interesting的点的话，就要换了;问最长能看多上时间;

思路：

那就看相邻的点的间隔是不是长于15了,还有就是这个节目一共长90分钟;

AC代码：

#include <bits/stdc++.h>

using namespace std;

#define Riep(n) for(int i=1;i<=n;i++)

#define Riop(n) for(int i=0;i<n;i++)

#define Rjep(n) for(int j=1;j<=n;j++)

#define Rjop(n) for(int j=0;j<n;j++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef long long LL;

const LL mod=1e9+;

const double PI=acos(-1.0);

const int inf=0x3f3f3f3f;

const int N=1e5+;

int n,a[];

int main()

{

    scanf("%d",&n);

    for(int i=;i<=n;i++)

    {

        scanf("%d",&a[i]);

    }

    a[]=;

    for(int i=;i<=n;i++)

    {

        if(a[i]-a[i-]>)

        {

            cout<<a[i-]+<<endl;

            return ;

        }

    }

    cout<<min(,a[n]+)<<endl;

    return ;

}