最后$f(p^x,p^y)$为求x,y的最大公约数，但返回的结果为p^gcd(x,y)
辗转相除法 o(logn) (a,b) = (b,a%b)
辗转相减法 o(n) (a,b) = (b,a-b)

https://blog.csdn.net/destiny_balabala/article/details/105074056

两个指数相减可以转化为相除
两个指数相除开根号

IA = lambda:map(int,input().split())
x = list(IA())
n = x[0]
if len(x)>1:
    a = x[1:]
else:
    a = list(IA())
res = [[0,0] for i in range(len(a)-1)]
a = list(set(a))
a = sorted(a)

def gcd(a, b):
    if b != 0:
        return gcd(b, a%b)
    return a
def gcd_sub(a, b):
    if a < b:a,b = b,a
    if b==1:
        return a
    return gcd_sub(b, a//b)

for i in range(1,len(a)):
    temp = gcd(a[i],a[0])
    res[i-1][0] = a[i] // temp
    res[i-1][1] = a[0] // temp


up = res[0][0]
down = res[0][1]

for i in range(1, len(a)-1):
    up = gcd_sub(up, res[i][0])
    down = gcd_sub(down,res[i][1])
print(str(up)+"/"+str(down))