Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … 
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ? 

Input

The first line contain a integer T , the number of cases. 
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2 31).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<string>
using namespace std;
typedef long long ll;
const int maxn = 10010;
const int maxv = 10010;
ll w[maxn],c[maxn],dp[maxv];
int main(){
    int t;
    cin>>t;
    while(t--)
    {
        memset(w,0,sizeof(w));
        memset(c,0,sizeof(c));
        memset(dp,0,sizeof(dp));
        int n,V;
        cin>>n>>V;
        for(int i = 1; i <= n; i++)
        {
            cin>>c[i];//價值
        }
        for(int i = 1; i <= n ;i++)
        {
            cin>>w[i];//重量
        }
        //邊界
        for(int v = 0; v <=V; v++)
        {
            dp[v] = 0;
        }
        for(int i = 1; i <=n; i++)
        {
            for(int v=V; v>=w[i];v--){
                dp[v] = max(dp[v],dp[v-w[i]]+c[i]);
            }
        }
        ll res = 0;
        for(int v = 0;v<=V;v++)
        {
            if(res<dp[v]){
                res = dp[v];
            }
        }
        cout<<res<<endl;
    }
}