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;
vector<long long int> factors(long long int number) {
  vector<long long int> result;
  for (long long int f = 2; f * f <= number; f++) {
    if (number % f != 0) continue;
    long long int factor = 1;
    while (number % f == 0) {
      factor *= f;
      number /= f;
    }
    result.push_back(factor);
  }
  if (number != 1) result.push_back(number);
  return result;
}
int main() {
  long long int X;
  cin >> X;
  vector<long long int> facts = factors(X);
  int n = facts.size();
  long long int minmax = X + 1;
  long long int bestProduct = -1;
  long long int bestNotProduct = -1;
  for (int bitset = 0; bitset < (1 << n); bitset++) {
    long long int product = 1;
    long long int notProduct = 1;
    for (int i = 0; i < n; i++) {
      if (bitset & (1 << i)) {
        product *= facts[i];
      } else
        notProduct *= facts[i];
    }
    if (minmax > max(product, notProduct)) {
      minmax = max(product, notProduct);
      bestProduct = product;
      bestNotProduct = notProduct;
    }
  }
  cout << bestProduct << " " << bestNotProduct << endl;
  return 0;
}