/* A Naive recursive implementation of
0-1 Knapsack problem */
#include <bits/stdc++.h>
using namespace std;

// A utility function that returns
// maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }

// Returns the maximum value that
// can be put in a knapsack of capacity W
int knapSack(int W, vector<int> &wt, vector<int> val, int N)
{

    // Base Case
    if (n == 0 || W == 0)
        return 0;

    // If weight of the nth item is more
    // than Knapsack capacity W, then
    // this item cannot be included
    // in the optimal solution
    if (wt[n - 1] > W)
        return knapSack(W, wt, val, n - 1);

    // Return the maximum of two cases:
    // (1) nth item included
    // (2) not included
    else
        return max(
            val[n - 1]
                + knapSack(W - wt[n - 1],
                        wt, val, n - 1),
            knapSack(W, wt, val, n - 1));
}

// Driver code
int main()
{
    int N,W;
    cin>>N>>W;
    vector<int>wt(N),val(N);
    for(int i=0;i<N;i++)
    {
        cin>>wt[i];
    }
    for(int i=0;i<N;i++)
    {
        cin>>val[i];
    }
    cout << knapSack(W, wt, val, N);
    return 0;
}

// This code is contributed by rathbhupendra