Codeforces 439 A. Devu, the Singer and Churu, the Joker

Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.

Devu has provided organizers a list of the songs and required time for singing them. He will sing n songs, ith song will take ti minutes exactly.

The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.

People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.

You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:

The duration of the event must be no more than d minutes;
Devu must complete all his songs;
With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.

If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.

Input

The first line contains two space separated integers n, d (1 ≤ n ≤ 100; 1 ≤ d ≤ 10000). The second line contains n space-separated integers: t1, t2, ..., tn (1 ≤ ti ≤ 100).

Output

If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.

Sample test(s)

input

3 30
2 2 1

output

input

3 20
2 1 1

output

-1

Note

Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:

First Churu cracks a joke in 5 minutes.
Then Devu performs the first song for 2 minutes.
Then Churu cracks 2 jokes in 10 minutes.
Now Devu performs second song for 2 minutes.
Then Churu cracks 2 jokes in 10 minutes.
Now finally Devu will perform his last song in 1 minutes.

Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.

Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.

关于这题的解题报告：本人觉得当你想多了的时候就会把太多的时间浪费掉了，因为本人就体验过，算了发了一下牢骚就打住吧；

对于这题：基本思路就是简单的模拟；

下面就是本人的代码：

 1 #include<cstdio>
 2 #define N 100
 3 
 4 int t[N];
 5 int main()
 6 {
 7     int d,n;
 8     int Sum = 0;
 9     int count = 0;
10     
11     scanf("%d%d",&n,&d);
12     for(int i=1;i<=n;++i){
13         scanf("%d",&t[i]);
14         Sum += t[i];
15     }
16     Sum += (n-1)*10;
17     if(Sum > d)
18          printf("-1
");
19     else
20     {
21         count += (n-1)*2;
22         printf("%d
",count+(d-Sum)/5);
23     }
24     int a;scanf("%d
",&a);
25     return 0;
26 }