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:
from collections import Counter
import string
import math
import sys
def array_int():
    return [int(i) for i in sys.stdin.readline().split()]
def vary(number_of_variables):
    if number_of_variables==1:
        return int(sys.stdin.readline())
    if number_of_variables>=2:
        return map(int,sys.stdin.readline().split()) 
def makedict(var):
    return dict(Counter(var))
mod=100000007
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
def printDivisors(n) :
    divisors=[]
    # Note that this loop runs till square root
    i = 1
    while i <= math.sqrt(n):
         
        if (n % i == 0) :
             
            # If divisors are equal, print only one
            if (n//i == i) :
                divisors.append(i)
            else :
                # Otherwise print both
                divisors.extend((i,n//i))
        i = i + 1
    return divisors
n=vary(1)
z=printDivisors(n)
if len((z))==2 or len(z)==1:
    print(1,n)
else:
    ans=float('inf')
    for i in range(len(z)):
        if (z[i]*(n//z[i]))//math.gcd(z[i],(n//z[i]))==n:
            ans=min(ans,max(z[i],n//z[i]))
    print(ans,n//ans)