Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.

LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer X (1 ≤ X ≤ 10^{12}).

Output

Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.

Examples

Input


2


Output


1 2


Input


6


Output


2 3


Input


4


Output


1 4


Input


1


Output


1 1
def div(x):
    a = dict()
    i = 2
    x_a = x
    while i * i <= x_a:
        if x % i == 0:
            if i in a:
                a[i] = a[i] * i
            else:
                a[i] = i
            x //= i
        else:
            i += 1
    if x != 1:
        if x in a:
            a[x] = a[x] * x
        else:
            a[x] = x
    return a


def check(i, a, x):
    b = bin(i)[2:]
    b = '0' * (len(a) - len(b)) + b
    ans = 1
    for i in range(len(a)):
        if b[i] == '1':
            ans *= a[i]
    return max(ans, x // ans)


x = int(input())
a = div(x)
ans = 10**13
b = []
for i in a:
    b.append(a[i])
for i in range(2**(len(a))):
    ans = min(ans, check(i, b, x))
print(x // ans, ans)