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 JAVA):

Solution:
import java.util.*;
public class water
{
    public static long min,minp,pit=0;
    public static long n;
    public static void dg(long  a[],int x,int y,int len,long sum1){
        for(int i=y;i<len-x+1;i++){
            Long sum=sum1;
            if(x==1){
                sum=sum*a[i];
                long mint=sum+n/sum;
               pit++;
                if(mint<min){
                    min=mint;
                    minp=sum;
                }
            }
            else{
                sum=sum*a[i];
                dg(a,x-1,i+1,len,sum);
            }
        }
        }
    public static void main(String args[])
    {
        Scanner ob=new Scanner(System.in);
        n=ob.nextLong();
        long x=n;
        long ax []=new long[10000000];
        int j=0;
        ax[j]=1;
         for(long i=2;i<Math.sqrt(n)+1&&x!=1;i++){
            for(;;){
                 if(x%i==0){
                    x=x/i;
                    ax[j]=ax[j]*i;
                }
                else if(ax[j]!=1){
                    j++;
                    ax[j]=1;
                    break;
                }
                else
                    break;
            }
        }
        if(x!=1){
             ax[j]=x;
             j++;
        }
        if(j==1){
            min=n+1;
            minp=1;
        }

        else {
            min=n+1;
            minp=1;
            for(int i=1;i<=(j/2);i++){
                dg(ax,i,0,j,1);
            }
        }
       
        if(minp>n/minp){
            System.out.println(n/minp+" "+minp);
        }
        else{
            System.out.println(minp+" "+n/minp);
        }
    }}