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.math.BigInteger;
import java.util.*;
import java.io.*;

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); }
        String next() { while (st == null || !st.hasMoreElements()) {
            try { st = new StringTokenizer(br.readLine()); }
            catch (IOException e) { e.printStackTrace(); } }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }

        String nextLine() {
            String str = "";
            try { str = br.readLine(); }
            catch (IOException e) { e.printStackTrace(); }
            return str;
        }
    }

    static long helper(long a,long b){
        if(b==0){
            return a;
        }else{
            return helper(b,a%b);
        }
    }
    public static void main(String[] args) throws IOException {
        FastReader ip = new FastReader();
        OutputStream output = System.out;
        PrintWriter out = new PrintWriter(output);
        long n=ip.nextLong();
        long val=(long)Math.sqrt(n);
        long a1=0;
        long a2=0;
        while (val>=1){
            if(n%val==0){
                if(helper(val,n/val)==1){
                    a1=val;
                    a2=n/val;
                    break;
                }
            }
            val--;
        }
        out.println(a1+" "+a2);
        out.close();
    }
}