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;
template <class T>
inline void read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    gcd(b, a % b);
}
int main() {
  long long n;
  read(n);
  long long x = sqrt(n);
  if (x * x == n) x -= 1;
  for (long long i = x; i >= 1; i--) {
    if (n % i == 0) {
      if (gcd(n / i, i) == 1) {
        printf("%lld %lld", i, n / i);
        return 0;
      }
    }
  }
  printf("1 %lld", n);
  return 0;
}