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;
long long GCD(long long a, long long b) {
  if (b)
    while ((a %= b) && (b %= a))
      ;
  return a + b;
}
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
int main() {
  long long X;
  cin >> X;
  long long res = X;
  long long r1 = 1, r2 = X;
  for (long long i = 1; i * i <= X; i++) {
    if (X % i == 0) {
      long long j = X / i;
      if (LCM(i, j) == X && res > max(i, j)) {
        res = max(i, j);
        r1 = i;
        r2 = j;
      }
    }
  }
  cout << r1 << " " << r2 << endl;
  return 0;
}
