// O(A^2) DP解法
// 入力制約: 0 <= n <= 10^13, 0 < a_i < 2048, 0 <= b_i, b_max * floor(n / a_min) <= 2^61
fn solve(n: u64, mut ab: [(u64, u64); 3]) -> u64 {
    // 割引額/使用枚数 の効率が最大の組を (a1,b1) に、それ以外は (a2,b2),(a3,b3) に
	if ab[1].0 * ab[0].1 < ab[0].0 * ab[1].1 {
		ab.swap(0, 1);
	}
	if ab[2].0 * ab[0].1 < ab[0].0 * ab[2].1 {
		ab.swap(0, 2);
	}
	let [(a1, b1), (a2, b2), (a3, b3)] = ab;
	// w: 確定で(a1,b1)の割引を使う回数
	let w = (n / a1).saturating_sub(a2 + a3);
	// m: DPで調べる分の「うどん札」の残り枚数 
	let m = n - w * a1;
	debug_assert_eq!((a1 * (a2 + a3) + (n % a1)).min(n), m);
	let mut dp = vec![0u64; (m + a1.max(a2).max(a3) + 1) as usize];
	// (w * b1): 確定で(a1,b1)の割引を使った分の割引合計額
	let mut r = w * b1;
	// 配るDP
	for i in 0..=m {
		dp[i as usize].chmax(r);
		r = dp[i as usize];
		dp[(i + a1) as usize].chmax(r + b1);
		dp[(i + a2) as usize].chmax(r + b2);
		dp[(i + a3) as usize].chmax(r + b3);
	}
	r
}

fn main() {
    let mut stdinlock = std::io::stdin().lock();
    let mut lines = std::io::BufRead::lines(&mut stdinlock).map_while(Result::ok);
	let n = lines.next().unwrap().parse::<u64>().unwrap();
    let mut ab = [(0u64, 0u64); 3];
    for e in ab.iter_mut() {
		let s = lines.next().unwrap();
		let mut t = s.split_whitespace();
		*e = (
            t.next().unwrap().parse::<u64>().unwrap(),
            t.next().unwrap().parse::<u64>().unwrap(),
		);
    }
    println!("{}", solve(n, ab));
}

trait Change {
    fn chmax(&mut self, x: Self);
    fn chmin(&mut self, x: Self);
}
impl<T: PartialOrd> Change for T {
    fn chmax(&mut self, x: T) {
        if *self < x {
            *self = x;
        }
    }
    fn chmin(&mut self, x: T) {
        if *self > x {
            *self = x;
        }
    }
}