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 Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        long n = scanner.nextLong();
        HashMap<Long, Long> map = new HashMap<>();
        primeFactors(n, map);
        if (map.size() == 1 || map.size() == 0) {
            System.out.println(n + " " + 1);
        } else {
            List<Long> factors = new ArrayList<>();
            for (Map.Entry<Long , Long> entry : map.entrySet()) {
                factors.add((long) Math.pow(entry.getKey(), entry.getValue()));
            }
            Collections.sort(factors);
            long a = factors.get(0);
            long b = factors.get(factors.size()-1);
            for (int i = 1; i < factors.size()-1; i++) {
                if (a < b) {
                    a*=factors.get(i);
                } else {
                    b*=factors.get(i);
                }
            }
            long ans = solve(1,1,0, factors);
            System.out.println(ans + " " + n/ans);
        }
    }

    private static long solve(long l, long r, int idx, List<Long> arr) {
        if (idx == arr.size()-1) {
            return Math.min(Math.max(l*arr.get(idx), r), Math.max(l, arr.get(idx)*r));
        }
        return Math.min(solve(l*arr.get(idx), r, idx+1, arr), solve(l, r*arr.get(idx), idx+1, arr));
    }

    public static void primeFactors(long n, HashMap<Long, Long> map) {
        while (n % 2 == 0) {
            map.putIfAbsent((long)2, (long)0);
            map.put((long)2, map.get((long)2) + 1);
            n /= 2;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            while (n % i == 0) {
                map.putIfAbsent((long)i, (long)0);
                map.put((long)i, map.get((long)i)+1);
                n /= i;
            }
        }

        if (n > 2)
            map.putIfAbsent(n, (long)1);
    }
}