In a certain course, you take n tests. If you get ai out of bi 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 ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 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 1

5 0 2

5 1 6

4 2

1 2 7 9

5 6 7 9

0 0

Sample Output

83

100

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：

n场考试中分别答对a_i题，总题数分别为b_i，允许翘掉k场考试，求能达到的最高准确率。

题解：

二分查找准确率即可。（这类题贪心肯定不行，通过样例也可以看出来）。

题目测试数据可以在这里找：

#include
    
     #include
     
      #include
      
       #include
       
        using namespace std;const int maxn=1e5+5;const double INF=1e6+5,esp=1e-6;int v[maxn],w[maxn];double y[maxn];int n,k;bool check(double x)//可以选择使得单位重量的价值不小于x{    for(int i=0;i
        
         =0;}void solve(){    double lb=0,ub=INF;    while(lb+esp
         
          >n>>k,n||k) { for(int i=0;i<2*n;i++) { if(i