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;
int prime(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return 0;
    }
  }
  return 1;
}
int gcd(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test, l = 0, g = 0, a = 0, b = 0;
  cin >> test;
  long long mini = LONG_LONG_MAX;
  if (prime(test)) {
    cout << 1 << " " << test << endl;
  } else {
    for (long long i = 1; i * i <= test; i++) {
      if (test % i == 0) {
        if (gcd(i, test / i) == 1) {
          g = max(i, test / i);
          if (g <= mini) {
            mini = g;
          } else {
            break;
          }
        }
      }
    }
    cout << mini << " " << test / mini << endl;
  }
  return 0;
}