#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define fi first #define se second #define rep(i,n) for(int i = 0; i < n; ++i) #define rrep(i,n) for(int i = 1; i <= n; ++i) #define drep(i,n) for(int i = n-1; i >= 0; --i) #define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next) #define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) #define rng(a) a.begin(),a.end() #define maxs(x,y) x = max(x,y) #define mins(x,y) x = min(x,y) #define pb push_back #define sz(x) (int)(x).size() #define pcnt __builtin_popcount #define snuke srand((unsigned)clock()+(unsigned)time(NULL)); inline int in() { int x; scanf("%d",&x); return x;} using namespace std; typedef long long int ll; typedef pair P; typedef vector vi; const int MX = 100005, INF = 1000010000; const ll LINF = 1000000000000000000ll; const double eps = 1e-10; const int di[] = {-1,0,1,0}, dj[] = {0,-1,0,1}; //^ map dp; int d[2]; vi a, b[2]; int sol(vi a) { if (dp.count(a)) return dp[a]; int res = 1; rep(k,2) { int i = min(a[0],d[k]/1000); int r = d[k]-1000*i; int j = min(a[1],r/100); r -= 100*j; if (r > a[2]) continue; vi x = a; x[0] -= i; x[1] -= j; x[2] -= r; rep(l,3) x[l] += b[k][l]; res = max(res,sol(x)+1); } return dp[a] = res; } int main(){ a = b[0] = b[1] = vi(3); rep(i,3) cin >> a[i]; cin >> d[0]; rep(i,3) cin >> b[0][i]; cin >> d[1]; rep(i,3) cin >> b[1][i]; cout << sol(a)-1 << endl; return 0; }