# 包除原理、ビット全探索
# Algo X Math ID068

N = int(input())
abc = list(map(int, input().split()))
K = 3

from math import gcd

ans = 0 
for bit in range(1<<K):
    if bit == 0:
        continue
    count = 0
    LCM = 1
    for shift in range(K):
        if bit>>shift & 1 == 1:
            count += 1
            LCM = LCM*abc[shift]//gcd(LCM, abc[shift])
    if count%2 == 1:
        ans += N//LCM
    else:
        ans -= N//LCM
    #print(bin(bit), count, LCM, N//LCM, ans)

print(ans)