Problem: 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

Solution (in PYTHON3):

Solution:
def getPrimes(n):
    isPrime = [True] * n
    p = []
    for i in range(2, n):
        if isPrime[i]:
            for j in range(i, (n - 1) // i + 1):
                isPrime[i * j] = False
            p.append(i)
    return p

def main():
    n = int(input())

    p = getPrimes(int(n ** 0.5 + 0.5) + 100)

    a = []

    for x in p:
        if n % x == 0:
            a.append([x, 0])
            while n % x == 0:
                n //= x
                a[-1][-1] += 1
    if n > 1 or len(a) == 0:
        a.append([n, 1])

    a = [x[0] ** x[1] for x in a]
    ans = solve(a)
    print(ans[0], ans[1])

def solve(a):
    n = len(a)
    ans = (float('inf'), (0, 0))
    for x in range(1<<n):
        s0, s1 = 1, 1
        for i in range(n):
            if (x & (1 << i)):
                s1 *= a[i]
            else:
                s0 *= a[i]
        ans = min(ans, (max(s0, s1), (s0, s1)))
    return ans[1]
main()