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:
## implementation 1 using all possible combinations
def primes(n):
    dick = {}
    while(n%2 == 0):
        dick[2] = dick.get(2, 0) + 1
        n = n // 2
    for i in range(3, int(n**0.5)+1, 2):
        while(n%i == 0):
            dick[i] = dick.get(i, 0) + 1
            n = n // i
    if n > 2:
        dick[n] = dick.get(n, 0) + 1
    return dick

def combs(this, index, curr):
    global ans
    
    if index == len(this):
        ans.append(curr)
        return
    combs(this, index+1, curr*this[index])
    combs(this, index+1, curr)

n = int(input())
dick = primes(n)
array = []
for i in dick:
    array.append(i**dick[i])

index = 0; curr = 1; ans = []
combs(array, index, curr)
answer = 1e13
for i in ans:
    answer = min(answer, max(i, n//i))
print(min(answer, n//answer), max(answer, n//answer))