#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int N = 60;

// 价格：i元，最后两次分别编号为j，k的最大长度和
int n, c, li[N], wi[N], ans, dp[N][N][N];

int main() {
    scanf("%d%d", &n, &c);
    for (int i = 1; i <= n; ++i) scanf("%d", &li[i]);
    for (int i = 1; i <= n; ++i) scanf("%d", &wi[i]);

    memset(dp, 0xcf, sizeof(dp));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (li[i] == li[j] || wi[i] + wi[j] > c) continue;
            dp[wi[i] + wi[j]][i][j] = li[i] + li[j];
        }
    }

    for (int i = 0; i <= c; ++i) {
        for (int j = 1; j <= n; ++j) {
            for (int k = 1; k <= n; ++k) {
                if (li[j] == li[k]) continue;
                for (int l = 1; l <= n; ++l) {
                    if (li[k] == li[l]) continue;
                    if ((li[j] < li[k] && li[k] > li[l] && li[j] != li[l])
                            || (li[j] > li[k] && li[k] < li[l] && li[j] != li[l])) {
                        if (i + wi[l] > c || dp[i][j][k] < 0) continue;
                        dp[i + wi[l]][k][l] = max(dp[i + wi[l]][k][l], dp[i][j][k] + li[l]);
                        ans = max(ans, dp[i + wi[l]][k][l]);
                    }
                }
            }
        }
    }

    printf("%d\n", ans);

    return 0;
}