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;
long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long t = gcd(a, b);
  return (a * b) / t;
}
int main() {
  long long x, ma = 0, t;
  cin >> x;
  long long a, b, temp;
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      a = i;
      b = x / i;
      temp = lcm(a, b);
      if (temp == x) {
        ma = max(a, ma);
      }
    }
  }
  printf("%lld %lld\n", ma, x / ma);
}