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
from sys  import stdin,stdout
 
import bisect
 
import math
 
def st():
    return list(stdin.readline().strip())
 
def inp():
    return int(stdin.readline())
 
def li():
    return list(map(int,stdin.readline().split()))
 
def mp():
    return map(int,stdin.readline().split())
 
def pr(n):
    stdout.write(str(n)+"\n")
 
def soe(limit):
    l=[1]*(limit+1)
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
 
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime
 
def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
 
def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r
 
def extgcd(a,b):
    if a==0:
        return b,0,1

    g,x1,y1=extgcd(b%a,a)
    x=y1-(b//a)*x1
    y=x1
    return g,x,y

def solve():
    n=inp()
    i=1
    while i*i<=n:
        if n%i==0 and (i*(n//i))//math.gcd(i,n//i)==n:
            c=i
        i+=1
    print(c,n//c)

   
                
                
            
 
for _ in range(1):
    solve()
