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 java.util.*;
import java.io.*;
public class Main
{
    public static void main(String []args) throws IOException {
    	FastScanner in  = new FastScanner(System.in);
    	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
    	solve(in, out);
    	in.close();
    	out.close();
    }
    static long gcd(long a, long b){ return (b==0) ? a : gcd(b, a%b); }
    static int gcd(int a, int b){ return (b==0) ? a : gcd(b, a%b); }
    static int lcm(int a, int b){ return (int)((a*1L*b)/(gcd(a,b)+0L)); }
    static long lcm(long a, long b){ return (a*b)/(gcd(a,b)+0L); }
    private static int fact(int n) { int ans=1; for(int i=2;i<=n;i++) ans*=i; return ans; }
    
    public static int[] generatePrimes(int n) {
		boolean[] prime = new boolean[n + 1];
		Arrays.fill(prime, 2, n + 1, true);
		for (int i = 2; i * i <= n; i++)
			if (prime[i])
				for (int j = i * i; j <= n; j += i)
					prime[j] = false;
		int[] primes = new int[n + 1];
		int cnt = 0;
		for (int i = 0; i < prime.length; i++)
			if (prime[i])
				primes[cnt++] = i;
		return Arrays.copyOf(primes, cnt);
	}
    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}
	String next(){while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}
    }
    
    /********* SOLUTION STARTS HERE ************/
    static long ans=0L,M=(long)1e9+7;
    private static void solve(FastScanner in, PrintWriter out){
        long a = 0L, b = 0L;
        long n = in.nextLong();
        for(long i=1;i*i<=n;i++){
            if(n%i==0){
                if(lcm(i,n/i)==n){
                    a = i;
                    b = n/i;
                }
            }
        }
        out.println(a+" "+b);
    }
    
    /************* SOLUTION ENDS HERE **********/
}