// ======================== ORIGINAL PYTHON CODE (AS REQUIRED) ========================
// from heapq import heappop, heappush, heapify
// from bisect import bisect
// #from sortedcontainers import SortedList
// from collections import deque, defaultdict
// from math import floor, ceil, isqrt, comb
// from sys import stdin, setrecursionlimit
// #setrecursionlimit(10**7)
// intin = lambda: int(stdin.readline())
// strin = lambda: stdin.readline().rstrip()
// listin = lambda: list(map(int, stdin.readline().split()))
// tuplein = lambda m: [tuple(map(lambda x: int(x) if x.isdigit() or (len(x) > 1 and x[0] == "-" and x[1:].isdigit()) else x, stdin.readline().split())) for _ in range(m)]
// gridin = lambda m: [list(map(int, stdin.readline().split())) for _ in range(m)]
// strgridin = lambda h: [stdin.readline().rstrip() for _ in range(h)]
// mapin = lambda: map(int, stdin.readline().split())
// N, D, K = mapin()
// A = listin()
// C = listin()
// dp = [[-float("inf")] * (K + 1) for _ in range(D + 1)]
// #dp[j][k] = i番目まで見たとき、jことったとき美しさがkであるときの満足度の最大値
// dp[0][0] = 0
// for i in range(N):
//     for j in range(D, -1, -1):
//         for k in range(K, -1, -1):
//             if j + 1 <= D:
//                 dp[j+1][min(K,k+C[i])] = max(dp[j+1][min(K,k+C[i])], dp[j][k] + A[i])
// print(dp[D][K] if dp[D][K] != -float("inf") else "No")
// ====================================================================================

#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, D, K;
    if (!(cin >> N >> D >> K)) return 0;
    vector<long long> A(N), C(N);
    for (int i = 0; i < N; ++i) cin >> A[i];
    for (int i = 0; i < N; ++i) cin >> C[i];

    const long long NEG = -(1LL << 60);
    // dp[j][k] = after processing some prefix of items,
    //            taking j items and beauty k, max satisfaction.
    vector<vector<long long>> dp(D + 1, vector<long long>(K + 1, NEG));
    dp[0][0] = 0;

    for (int i = 0; i < N; ++i) {
        for (int j = D; j >= 0; --j) {
            for (int k = K; k >= 0; --k) {
                if (j + 1 <= D && dp[j][k] != NEG) {
                    int nk = (int)min<long long>(K, k + C[i]);
                    dp[j + 1][nk] = max(dp[j + 1][nk], dp[j][k] + A[i]);
                }
            }
        }
    }

    long long ans = dp[D][K];
    if (ans == NEG) cout << "No\n";
    else cout << ans << '\n';
    return 0;
}