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
#include <bits/stdc++.h>
using namespace std;
void ini() {}
bool check(unsigned long long int n, unsigned long long int m,
           unsigned long long int x) {
  unsigned long long int res = n * m, temp;
  while (m != 0) {
    temp = m;
    m = n % m;
    n = temp;
  }
  if ((res / n) == x) {
    return true;
  } else {
    return false;
  }
}
int main() {
  ini();
  unsigned long long int x, d;
  cin >> x;
  d = sqrt(x);
  unsigned long long int a = 2, temp, m = x;
  while (a <= d) {
    if (x % a == 0 && check(x / a, a, x)) {
      temp = max(x / a, a);
      m = min(m, temp);
    }
    a++;
  }
  cout << m << " " << x / m;
}
