#include #include typedef long long int ll; using namespace std; typedef pair P; using namespace atcoder; template using min_priority_queue = priority_queue, greater>; #define USE998244353 #ifdef USE998244353 const ll MOD = 998244353; using mint = modint998244353; #else const ll MOD = 1000000007; using mint = modint1000000007; #endif #pragma region //使いがち const int MAX = 2000001; long long fac[MAX], finv[MAX], inv[MAX]; void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll pow_ll(ll x, ll n) { if (n == 0) return 1; if (n % 2) { return pow_ll(x, n - 1) * x; } else { ll tmp = pow_ll(x, n / 2); return tmp * tmp; } } // floor(a^(1/k)) ll floor_root(ll a, ll k) { assert(a >= 0); assert(k >= 1); if (a == 0) return 0; if (k == 1) return a; // 大体の値 ll x = (ll)pow(a, 1.0 / k); // 増やす while ((pow_ll(x + 1, k)) <= a) { x++; } // 減らす while ((pow_ll(x, k)) > a) { x--; } return x; } ll keta(ll num, ll arity) { ll ret = 0; while (num) { num /= arity; ret++; } return ret; } // k進数で見た時のi桁目の数を返す (一番下は0桁目) ll keta_num(ll num, ll i, ll k) { return (num / pow_ll(k, i)) % k; } ll ceil(ll n, ll m) { // n > 0, m > 0 ll ret = n / m; if (n % m) ret++; return ret; } void compress(vector& v) { // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] vector u = v; sort(u.begin(), u.end()); u.erase(unique(u.begin(),u.end()),u.end()); map mp; for (int i = 0; i < u.size(); i++) { mp[u[i]] = i; } for (int i = 0; i < v.size(); i++) { v[i] = mp[v[i]]; } } vector > prime_factorize(ll N) { vector > res; for (ll a = 2; a * a <= N; ++a) { if (N % a != 0) continue; ll ex = 0; // 指数 // 割れる限り割り続ける while (N % a == 0) { ++ex; N /= a; } // その結果を push res.push_back({a, ex}); } // 最後に残った数について if (N != 1) res.push_back({N, 1}); return res; } #pragma endregion // 1以上n以下について素数かどうかのリストを返す vector is_prime_list(int n) { vector is_prime(n + 1, true); is_prime[0] = false; is_prime[1] = false; for (int i = 2; i * i <= n; i++) { if (!is_prime[i]) continue; for (int j = i * 2; j <= n; j += i) { is_prime[j] = false; } } return is_prime; } int main() { int n; cin >> n; vector a(n); ll sum = 0; for (int i = 0; i < n; i++) { cin >> a[i]; sum += a[i]; } vector dp(850001, 0); dp[0] = 1; for (int i = 0; i < n; i++) { vector dp2(850001, 0); for (int j = 0; j < 850001; j++) { if (j + a[i] < 850001) { dp2[j + a[i]] += dp[j]; } } for (int j = 0; j < 850001; j++) { dp[j] += dp2[j]; } } for (ll i = 1; i < 850001; i++) { if ((dp[i] >= 2) && (i * 2 < sum)) { cout << sum - 2 * i << endl; exit(0); } } cout << -1 << endl; return 0; }