#include <bits/stdc++.h>
#include <atcoder/all>
 
#define elif else if
#define ll long long
#define vll vector<long long>
#define vec vector
#define embk emplace_back
#define rep(i, n) for (ll i = 0; i < n; i++)
#define rep3(i, n, k) for (ll i = k; i < n; i++)
#define all(a) a.begin(), a.end()
#define YNeos(bool) (bool ? "Yes" : "No")
 
using namespace std;
using namespace atcoder;
const ll INF = 1LL << 60;
const ll mod = 998244353;
const double pi = acos(-1);


int main() {
  ll n; cin >> n;
  vll a(3), b(3); rep(i, 3) cin >> a[i] >> b[i];

  ll mima = min(a[0], min(a[1], a[2]));
  ll minb = min(b[0], min(b[1], b[2]));

  //うどん札1枚当たりの割引額が最も大きいものを(a[0], b[0])にする
  if (a[1] * b[0] < a[0] * b[1]) {
    swap(a[0], a[1]);
    swap(b[0], b[1]);
  }
  if (a[2] * b[0] < a[0] * b[2]) {
    swap(a[0], a[2]);
    swap(b[0], b[2]);
  }
  
  //(a[0], b[0])の割引を必ず使う回数と、DPで調べるうどん札の残り枚数を求める
  ll use_0 = max(n / a[0] - a[1] - a[2], 0LL);
  ll nokori_n = min(a[0] * (a[1] + a[2]) + (n % a[0]), n);

  //DP
  vll dp(nokori_n + 2001);
  ll dpans = 0;

  rep(i, nokori_n + 1) {
    dp[i + a[0]] = max(dp[i] + b[0], dp[i + a[0]]);
    dp[i + a[1]] = max(dp[i] + b[1], dp[i + a[1]]);
    dp[i + a[2]] = max(dp[i] + b[2], dp[i + a[2]]);
    dpans = max(dp[i], dpans);
  }

  cout << dpans + use_0 * b[0] << endl;
}