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;
long long int gcd(long long int x, long long int y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int x;
  cin >> x;
  long long int a1 = 1, a2 = x;
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      long long int d = x / i;
      if (gcd(i, d) == 1) {
        long long int k = max(i, d);
        if (k < a2) {
          a2 = max(i, d);
          a1 = min(i, d);
        }
      }
    }
  }
  cout << a1 << " " << a2;
}