In a certain course, you take n tests. If you get a_i out of b_i questions correct on test i, your cumulative average is defined to be

.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

 #include<cstdio>

 #include<iostream>

 #include<algorithm>

 #include<cstring>

 #include<cmath>

 using namespace std;

 const int MAXN=;

 struct Node{

     int a,b;

 };

 Node dt[MAXN];

 double d[MAXN];

 int n,k;

 bool fsgh(double R){

     double sum=;

     for(int i=;i<n;i++)d[i]=dt[i].a-R*dt[i].b;

     sort(d,d+n);

     for(int i=n-;i>=n-k;i--)sum+=d[i];

     return sum>?true:false;

 }

 double erfen(double l,double r){

     double mid;

     while(r-l>1e-){

         mid=(l+r)/;

         if(fsgh(mid))l=mid;

         else r=mid;

     }

     return mid;

 }

 int main(){

     while(scanf("%d%d",&n,&k),n|k){

         double mx=;

         k=n-k;

         for(int i=;i<n;i++)scanf("%d",&dt[i].a);

         for(int i=;i<n;i++)scanf("%d",&dt[i].b),mx=max(1.0*dt[i].a/dt[i].b,mx);

         printf("%.0f\n",erfen(,mx)*);

     }

     return ;

 }