d/ F - Dropping test.（ai+加和 bi+ n k 求最小分数）

加和二分

为啥这样算

n 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_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.

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

#include 
#include
#include
using namespace std;
const int maxn=1000+5;
const int INF=1e8+10;
int n,k;
double a[maxn],b[maxn],c[maxn];
bool judge(double mid)
{
    for(int i=0;i());
    double ans=0;
    for(int i=0;i=0;
}
int main()
{
    while(cin>>n>>k&&n)
    {
        k=n-k;
        for(int i=0;i>a[i];
            a[i]=a[i]*100;
        }
        for(int i=0;i>b[i];
        double l=0,r=100.0;
        while(r-l>0.000001)
        {
            double mid=(l+r)/2.0;
            if(judge(mid)) l=mid;
            else
                r=mid;
        }
        int ans=(int )(l+0.5);
        cout << ans<< endl;
    }

    return 0;
}