#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, static_modint& a) {long long x; is >> x; a = x; return is;} template istream& operator>>(istream& is, dynamic_modint& a) {long long x; is >> x; a = x; return is;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} const long long INF = 1001001001001001; int main(){ int N, D, K; cin >> N >> D >> K; vector A(N); vector C(N); cin >> A >> C; vector> dp(D + 1, vector(K + 1, -INF)); dp[0][0] = 0; rep(i, N){ for(int j = D - 1; j >= 0; j--){ for(int k = K; k >= 0; k--){ if(dp[j][k] == -INF) continue; chmax(dp[j + 1][min(k + C[i], K)], dp[j][k] + A[i]); } } } long long ans = dp[D][K]; if(ans == -INF) cout << "No\n"; else cout << ans << "\n"; return 0; }