// 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 {
	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 = max(floor(n / a1) - (a2 + a3), 0)
	let w = (n / a1).saturating_sub(a2 + a3);
	let mut dp = [i64::MIN; 2048];
	let mut v = (w * b1) as i64;
	for i in 0..=(n - w * a1) {
        v = v
            .max(dp[(i.wrapping_sub(a1) & 2047) as usize] + (b1 as i64))
            .max(dp[(i.wrapping_sub(a2) & 2047) as usize] + (b2 as i64))
            .max(dp[(i.wrapping_sub(a3) & 2047) as usize] + (b3 as i64));
        dp[(i & 2047) as usize] = v;
	}
	v as u64
}

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));
}