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>
const double pi = 3.14159265358979323846;
using namespace std;
long long fastexp(long long a, long long b) {
  long long answer = 1;
  long long temp = a;
  while (b > 0) {
    if (b & 1) answer *= (temp);
    temp *= temp;
    b >>= 1;
  }
  return answer;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x;
  cin >> x;
  ;
  long long answer = x;
  long long a = 1, b = x;
  for (long long i = 2; i * i <= (x); i++) {
    if (x % i == 0) {
      long long q = (x / i);
      if (gcd(q, i) == 1) {
        answer = min(answer, max(q, i));
        a = i;
        b = q;
      }
    }
  }
  cout << a << " " << b << endl;
  return 0;
}