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
#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int lcm(long long int a, long long int b) {
  long long int p = a, q = b;
  while (b > 0) {
    long long int temp = b;
    b = a % b;
    a = temp;
  }
  return (p * q) / a;
}
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = sqrt(n); i > 0; i--) {
    if (n % i == 0 && lcm(i, n / i) == n) {
      cout << i << " " << n / i << '\n';
      break;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
