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 long long int mxN = 1e12 + 1;
const long long int MOD = 1e9 + 7;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
void solve() {
  set<pair<long long int, long long int>> s;
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (lcm(i, n / i) == n) s.insert(make_pair(i, n / i));
    }
  }
  pair<long long int, long long int> mn(mxN, mxN);
  for (auto a : s) {
    long long int x = max(a.first, a.second);
    long long int y = max(mn.first, mn.second);
    if (x < y) {
      mn = a;
    }
  }
  cout << mn.first << " " << mn.second << "\n";
}
signed main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}