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:
a=int(input())
d=[]

import math

if a==1:
    print(1,1)
else:
    for i in range (1,int(a**0.5)+1):
        if a%i==0:
            if a/i ==i:
                d.append(i)
            else:
                d.append(i)
                d.append(int(a/i))
    d.sort()
    if len(d)%2!=0:
        for i in range(int(len(d)/2)):
            if (d[int(len(d)/2)-1-i]*d[int(len(d)/2)+1+i])/math.gcd(d[int(len(d)/2)-1-i],d[int(len(d)/2)+1+i]) ==a:
                
                print(d[int(len(d)/2)-1-i],d[int(len(d)/2)+1+i])
                break
    else:
        for i in range(int(len(d)/2)):
            if (d[int(len(d)/2)-1-i]*d[int(len(d)/2)+i])/math.gcd(d[int(len(d)/2)-1-i],d[int(len(d)/2)+i]) ==a:
                print(d[int(len(d)/2)-1-i],d[int(len(d)/2)+i])
                break