Description

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 .

Input

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_ifor 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.

Output

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.

Sample Input

3 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

Hint

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

题意：有N个考试，每个考试有ai和bi两个值，最后成绩由上面的公式求得。幸运的是，可以放弃K个科目，求最大化最后的成绩。

思路：这是一道简单的最大化平均值模板题，化简出（ai-mid*bi）>0,求n-k前项。

AC代码：

#include
         
          #define INF 0x3f3f3f3f#include
          
           using namespace std;bool cmp(double x,double y){    return x>y;}int n,k;double y[1100];int a[1100];int b[1100];bool C(double  mid){   for(int i=0 ; i
           
            =0)    return true;   return false;}int main(){    while(scanf("%d%d",&n,&k)!=EOF)    {        if(n==0&&k==0)            break;        for(int i=0 ; i