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 (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long ans = (a * b) / gcd(a, b);
  return ans;
}
int main() {
  long long x;
  cin >> x;
  vector<long long> v;
  for (long long i = 1; i <= (long long)sqrt(x); i++) {
    if (x % i == 0) {
      v.push_back(i);
    }
  }
  long long minv = x;
  for (auto i = v.begin(); i != v.end(); i++) {
    if (lcm(*i, x / (*i)) == x) {
      long long temp = max(*i, x / (*i));
      minv = min(temp, minv);
    }
  }
  cout << minv << " " << x / minv << endl;
  return 0;
}