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：n个有趣时间点 当持续15分钟没有遇到有趣时间点 则截至  输出持续的总时间

题解：记录有趣时间点  for循环寻找 截至点并且输出   未找到则输出90

 #include<bits/stdc++.h>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<queue>

 #include<stack>

 #include<map>

 #define ll __int64

 #define pi acos(-1.0)

 using namespace std;

 int n;

 map<int,int> mp;

 int exm;

 int ans;

 int main()

 {

     int n;

     scanf("%d",&n);

     mp.clear();

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

     {

         scanf("%d",&exm);

         mp[exm]=;

     }

     int gg=;

     ans=;

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

     {

         if(mp[i])

         gg=i;

         if(i-gg>)

         {

             ans=i;

             break;

         }

     }

     if(ans==)

     cout<<""<<endl;

     else

     cout<<ans<<endl;

     return ;

 }