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
import math
import bisect
from collections import defaultdict
from collections import OrderedDict
from itertools import combinations 
#for _ in range(int(input())):
 
 
#a,b,c = map(int,input().split())
n = int(input())
#ar = list(map(int,input().split()))

#s  = list(input())

def factors(x):
    res = []
    i = 1
    while i*i <= x:

        if x%i == 0:
            res.append(i)

            if x//i != i:
                res.append(x//i)

        i += 1

    return res


fac = factors(n)
ans = 1e13
#print(fac)
#fac.sort()
#print(fac[0],fac[1])
for i in range(len(fac)):
    

    if (n//(math.gcd(fac[i],n//fac[i]))) == n:
        z = max(fac[i],n//fac[i])
        if z <= ans:
            a = fac[i]
            b = n//fac[i]
            ans = z

print(a,b)
                

        

    

        
        
        
        

    

            
    
        
    
    
    
 
    
 
 
"""
stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?)
    * do smth instead of nothing and stay organized
    * WRITE STUFF DOWN
    * DON'T GET STUCK ON ONE APPROACH
"""
    
