#!/usr/bin/env python3
# Based on http://pakapa104.hatenablog.com/entry/2016/07/02/014219

import array


INF = 10 ** 9


def sieve_of_eratosthenes(end):
    is_prime = array.array("B", (True for i in range(end)))
    is_prime[0] = False
    is_prime[1] = False
    for i in range(2, end):
        if is_prime[i]:
            for j in range(2 * i, end, i):
                is_prime[j] = False
    return is_prime


def compute_max_cups(init_money, cups):
    # 個数制限のないナップザック問題をボトムアップの動的計画法で解く
    # dp[i] = （所持金がiになるまでに買えるカップ麺の数）
    dp = array.array("l", (-INF for _ in range(init_money + 1)))
    dp[init_money] = 0
    for cup in cups:
        for i in range(init_money - cup + 1)[::-1]:
            dp[i] = max(dp[i], dp[i + cup] + 1)
    # 素数表を得る
    is_prime = sieve_of_eratosthenes(init_money + 1)
    # 金欠チャンスの処理
    # その後は、買えるだけ買う
    return sum(dp[i] for i in range(1, init_money + 1)
               if dp[i] > 0 and is_prime[i]) + max(dp)


def main():
    m = int(input())
    _ = int(input())
    cs = array.array("L", map(int, input().split()))
    print(compute_max_cups(m, cs))


if __name__ == '__main__':
    main()