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>
int gcd(long long n, long long m) {
  long long temp;
  while (m && n) {
    temp = n % m;
    n = m;
    m = temp;
  }
  if (n == 1)
    return 1;
  else
    return 0;
}
int main() {
  long long n, ans = 0, d, i;
  int c = 0;
  scanf("%lld", &n);
  for (i = 2; 1LL * i * i <= n; i++) {
    if (n % i == 0) {
      d = n / i;
      if (i > ans && i != d && gcd(i, d)) ans = i;
    }
    if (c >= 2) break;
  }
  if (ans == 0)
    printf("1 %lld", n);
  else
    printf("%lld %lld", ans, n / ans);
  return 0;
}
