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.*;
import java.math.*;

public class HelloWorld {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long X = scanner.nextLong();
        int N = (int)1e6;
        ArrayList<Long> primes = new ArrayList<>();
        for(long i = 2; i <= (long)Math.ceil(Math.sqrt(X)); i++) {
            if(X % i == 0) {
                long cur = 1;
                while(X % i == 0) {
                    X /= i;
                    cur *= i;
                }
                primes.add(cur);
            }
        }
        if(X > 1) primes.add(X);
        long ansA = Long.MAX_VALUE;
        long ansB = Long.MAX_VALUE;
        for(int i = 0; i < (1 << primes.size()); i++) {
            long a = 1;
            long b = 1;
            for(int k = 0; k < primes.size(); k++) {
                if(((i >> k) & 1) == 1) a *= primes.get(k);
                else b *= primes.get(k);
            }
            if(Math.max(a, b) < Math.max(ansA, ansB)) {
                ansA = a;
                ansB = b;
            }
        }
        System.out.println(ansA + " " + ansB);
    }
}