#include #include using namespace std; #include using namespace atcoder; using ll = long long; using vll = vector; using vvll = vector; using vvvll = vector; using vvvvll = vector; using vvvvvll = vector; #define all(A) A.begin(),A.end() #define rep(i, n) for (ll i = 0; i < (ll) (n); i++) template bool chmax(T& p, T q, bool C = 1) { if (C == 0 && p == q) { return 1; } if (p < q) { p = q; return 1; } else { return 0; } } template bool chmin(T& p, T q, bool C = 1) { if (C == 0 && p == q) { return 1; } if (p > q) { p = q; return 1; } else { return 0; } } ll gcd(ll(a), ll(b)) { if (a == 0)return b; if (b == 0)return a; ll c = a; while (a % b != 0) { c = a % b; a = b; b = c; } return b; } ll sqrtz(ll N) { ll L = 0; ll R = sqrt(N) + 10000; while (abs(R - L) > 1) { ll mid = (R + L) / 2; if (mid * mid <= N)L = mid; else R = mid; } return L; } ll inva(ll N, ll M) { ll mod = M; ll a = N, b = mod, c = 1, d = 0; while (b > 0) { ll t = a / b; a -= t * b; swap(a, b); c -= t * d; swap(c, d); } c %= mod; if (c < 0)c += mod; return c; } ll modPow(long long a, long long n, long long p) { if (n == 0) return 1; // 0乗にも対応する場合 if (n == 1) return a % p; if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p; long long t = modPow(a, n / 2, p); return (t * t) % p; } using mint = modint998244353; using vm = vector; using vvm = vector; using vvvm = vector; vector fact, factinv, inv, factK; ll mod = 998244353; void prenCkModp(ll n) { // factK.resize(4*n+5); fact.resize(n + 5); factinv.resize(n + 5); inv.resize(n + 5); fact[0] = fact[1] = 1; factinv[0] = factinv[1] = 1; inv[1] = 1; for (ll i = 2; i < n + 5; i++) { fact[i] = (fact[i - 1] * i); inv[i] = (mod - ((inv[mod % i] * (mod / i)))); factinv[i] = (factinv[i - 1] * inv[i]); } // factK[0]=1; // for(ll i=1;i<4*n+5;i++){ // factK[i]=factK[i-1]*mint(K-i+1); // //K*(K-1)*...*(K-i+1); // } } mint nCk(ll n, ll k) { if (n < k || k < 0) return 0; return (fact[n] * ((factinv[k] * factinv[n - k]))); } // ll K; // mint nCkK(ll n,ll k){ // if(K>N>>M>>K; vvll DP(N+1,vll(K+1,0)); DP[0][0]=1; rep(i,N){ rep(j,M){ ll a; cin>>a; rep(k,K+1){ if(k+a<=K&&DP[i][k]==1)DP[i+1][k+a]=1; } } } ll an=-1; rep(k,K+1)if(DP[N][k]==1)an=K-k; cout<