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 x;
long long a, b;
long long mn = 1e12;
int solve(long long d, long long f, long long s) {
  long long d1 = d, f1 = f;
  long long e = d % f;
  d = f;
  f = e;
  while (e) {
    e = d % f;
    d = f;
    f = e;
  }
  if (f1 * d1 / d == s) {
    return 1;
  } else {
    return 0;
  }
}
int main() {
  cin >> x;
  for (long long i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (max(i, x / i) < mn && solve(x / i, i, x) == 1) {
        a = i;
        b = x / i;
        mn = max(i, x / i);
      }
    }
  }
  cout << a << ' ' << b;
  return 0;
}
