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;
const int maxx = 100009;
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long _lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void solve() {
  long long x;
  cin >> x;
  vector<long long> ans(2);
  ans[0] = 1;
  ans[1] = x;
  long long maxm = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      if (i * i == x or _lcm(i, x / i) != x) {
        continue;
      }
      long long a = i;
      long long b = x / i;
      if (a < b) {
        swap(a, b);
      }
      if (a < maxm) {
        maxm = a;
        ans[0] = a;
        ans[1] = b;
      }
    }
  }
  cout << ans[0] << " " << ans[1] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t-- > 0) {
    solve();
  }
  return 0;
}