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;
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a / gcd(a, b) * b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t;
  cin >> t;
  long long int sum = 0;
  if (t == 1)
    cout << "1"
         << " "
         << "1" << '\n';
  else {
    for (long long int i = 1; i * i < t; ++i) {
      if (t % i == 0 && lcm(i, t / i) == t) {
        sum = i;
      }
    }
    cout << sum << " " << t / sum << '\n';
  }
}