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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
unsigned long long int gcd(unsigned long long int a, unsigned long long int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
unsigned long long int lcm(unsigned long long int a, unsigned long long int b) {
  unsigned long long int x;
  x = (a * b) / gcd(a, b);
  return x;
}
int main() {
  unsigned long long int n, x, y;
  cin >> n;
  unsigned long long int i, a = 9999999999999;
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (max(i, n / i) <= a && lcm(i, n / i) == n) {
        a = max(i, n / i);
        x = min(i, n / i);
        y = n / x;
      }
    }
  }
  cout << x << " " << y;
  return 0;
}