#include using namespace std; typedef long long signed int LL; typedef long long unsigned int LU; #define incID(i, l, r) for(int i = (l) ; i < (r); i++) #define incII(i, l, r) for(int i = (l) ; i <= (r); i++) #define decID(i, l, r) for(int i = (r) - 1; i >= (l); i--) #define decII(i, l, r) for(int i = (r) ; i >= (l); i--) #define inc(i, n) incID(i, 0, n) #define inc1(i, n) incII(i, 1, n) #define dec(i, n) decID(i, 0, n) #define dec1(i, n) decII(i, 1, n) #define inII(v, l, r) ((l) <= (v) && (v) <= (r)) #define inID(v, l, r) ((l) <= (v) && (v) < (r)) #define PB push_back #define EB emplace_back #define MP make_pair #define FI first #define SE second #define PQ priority_queue #define ALL(v) v.begin(), v.end() #define RALL(v) v.rbegin(), v.rend() #define FOR(it, v) for(auto it = v.begin(); it != v.end(); ++it) #define RFOR(it, v) for(auto it = v.rbegin(); it != v.rend(); ++it) template bool setmin(T & a, T b) { if(b < a) { a = b; return true; } else { return false; } } template bool setmax(T & a, T b) { if(b > a) { a = b; return true; } else { return false; } } template bool setmineq(T & a, T b) { if(b <= a) { a = b; return true; } else { return false; } } template bool setmaxeq(T & a, T b) { if(b >= a) { a = b; return true; } else { return false; } } template T gcd(T a, T b) { return (b == 0 ? a : gcd(b, a % b)); } template T lcm(T a, T b) { return a / gcd(a, b) * b; } // ---- ---- int a[3][5], dp[10001][11][101]; int main() { // 0 1 2 3 4 // <1000, 100, 1, sum, d> inc(k, 3) { if(k != 0) { cin >> a[k][4]; } inc(i, 3) { cin >> a[k][i]; } a[k][3] = a[k][0] * 1000 + a[k][1] * 100 + a[k][2]; } int ans = 0; dp[ a[0][3] ][ a[0][0] ][ a[0][1] ] = 1; // +1 decII(i, 0, 10000) { decII(j, 0, 10) { decII(k, 0, 100) { if(dp[i][j][k] == 0) { continue; } setmax(ans, dp[i][j][k]); int l = i - (j * 1000 + k * 100); incII(m, 1, 2) { int d = a[m][4]; if(i < d) { continue; } d -= min(j, d / 1000) * 1000; d -= min(k, d / 100) * 100; d -= min(l, d / 1) * 1; if(d != 0) { continue; } d = a[m][4]; // 計算量が…… incII(jj, 0, j) { incII(kk, 0, k) { int ll = d - (jj * 1000 + kk * 100); if(! inII(ll, 0, l)) { continue; } setmax(dp[i - d + a[m][3]][j - jj + a[m][0]][k - kk + a[m][1]], dp[i][j][k] + 1); } } } } } } cout << (ans - 1) << endl; // -1 return 0; }