CodeForces 42A Guilty — to the kitchen!

Guilty — to the kitchen!

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 42A

Description

It‘s a very unfortunate day for Volodya today. He got bad mark in algebra and was therefore forced to do some work in the kitchen, namely to cook borscht (traditional Russian soup). This should also improve his algebra skills.

According to the borscht recipe it consists of n ingredients that have to be mixed in proportion  litres (thus, there should be a₁ ·x, ..., a_n ·x litres of corresponding ingredients mixed for some non-negative x). In the kitchen Volodya found out that he has b₁, ..., b_n litres of these ingredients at his disposal correspondingly. In order to correct his algebra mistakes he ought to cook as much soup as possible in a V litres volume pan (which means the amount of soup cooked can be between 0 and V litres). What is the volume of borscht Volodya will cook ultimately?

Input

The first line of the input contains two space-separated integers n and V (1 ≤ n ≤ 20, 1 ≤ V ≤ 10000). The next line contains n space-separated integers a_i (1 ≤ a_i ≤ 100). Finally, the last line contains n space-separated integers b_i (0 ≤ b_i ≤ 100).

Output

Your program should output just one real number — the volume of soup that Volodya will cook. Your answer must have a relative or absolute error less than 10^- 4.

Sample Input

Input

1 100140

Output

40.0

Input

2 1001 125 30

Output

50.0

Input

2 1001 160 60

Output

100.0

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 struct Node
 6 {
 7     double a;
 8     double b;
 9     double c;
10 }s[25];
11
12 bool cmp(Node x,Node y)
13 {
14     return x.c<y.c;
15 }
16 int main()
17 {
18     int n,i,j;
19     double a[25],b[25],c[25],v;
20     while(scanf("%d %lf",&n,&v)!=EOF)
21     {
22         for(i=1;i<=n;i++)
23             scanf("%lf",&s[i].a);
24         for(i=1;i<=n;i++)
25         {
26             scanf("%lf",&s[i].b);
27             s[i].c=s[i].b/s[i].a;
28         }
29         sort(s+1,s+n+1,cmp);
30         double ans=0;
31         for(i=1;i<=n;i++)
32             ans=ans+s[1].c*s[i].a;
33         if(ans>v)
34             ans=v;
35         printf("%lf\n",ans);
36     }
37     return 0;
38 }