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:
# Python program to find LCM of two numbers 

# Recursive function to return gcd of a and b 
def gcd(a,b): 
	if a == 0: 
		return b 
	return gcd(b % a, a) 

# Function to return LCM of two numbers 
def lcm(a,b): 
	return (a*b) / gcd(a,b) 



import math  
  
# method to print the divisors
f=[] 
def printDivisors(n) : 
      
    # 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) : 
                f.append(i) 
            else : 
                # Otherwise print both 
                f.append(i)
                f.append(n//i)
        i = i + 1

n=int(input())
printDivisors(n)
f.sort()
lll=len(f)

if len(f)%2==0:
    s1=lll//2-1
    s2=lll//2
else:
    s1=lll//2-1
    s2=lll//2+1
if n==1:
    print("1 1")
if (n!=1):
    while(True):
        if lcm(f[s1],f[s2])==n:
            print(f[s1],f[s2])
            break
        else:
            s1-=1
            s2+=1