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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      long long t = 1;
      while (n % i == 0) t *= i, n /= i;
      a.push_back(t);
    }
  if (n > 1) a.push_back(n);
  if (a.size() <= 1) a.push_back(1);
  if (a.size() <= 1) a.push_back(1);
  sort(a.begin(), a.end());
  pair<long long, long long> ans =
      pair<long long, long long>(LLONG_MAX, LLONG_MAX);
  for (int c = 0; c < (1 << a.size()); c++) {
    long long u = 1, v = 1;
    for (int i = 0; i <= a.size() - 1; i++)
      if ((c >> i) & 1)
        u *= a[i];
      else
        v *= a[i];
    if (max(ans.first, ans.second) > max(u, v))
      ans = pair<long long, long long>(u, v);
  }
  cout << ans.first << ' ' << ans.second;
  return 0;
}
