# n=100だから全探索O(n^3)

def gcd(a,b):
    while b:a,b=b,a%b
    return a

n=int(input())
aa=[int(input()) for _ in range(n)]

ans=0
for i,a3 in enumerate(aa):
    for j,a2 in enumerate(aa[:i]):
        g23=gcd(a2,a3)
        for a1 in aa[:j]:
            if gcd(g23,a1)==1:ans+=1

print(ans)