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>
const long long int N = 2e5 + 5;
using namespace std;
bool isPrime(long long int n) {
  if (n == 1) return false;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
void CSK() {
  long long int x;
  cin >> x;
  long long int as1 = 0, as2 = 0;
  for (long long int i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      if (lcm(i, x / i) == x) {
        as1 = i;
        as2 = x / i;
      }
    }
  }
  cout << as1 << " " << as2 << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cout.precision(10);
  long long int __TESTCASES = 1;
  while (__TESTCASES--) {
    CSK();
  }
}
