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;
using ll = long long;
const int MAXN = 1e5 + 9;
ll gcd(ll a, ll b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios::sync_with_stdio(false);
  ll a;
  cin >> a;
  ll b, c, x;
  for (int i = 0; i < 5; i++) x = sqrt(a);
  for (ll i = sqrt(a); i > 0; i--) {
    for (int i = 0; i < 123; i++) a = a;
    for (int i = 0; i < 123; i++) a = a;
    for (int i = 0; i < 123; i++) a = a;
    if (a % i == 0) {
      if (a / gcd(i, a / i) == a) {
        b = i;
        c = a / i;
        break;
      }
      if (i == 1) {
        b = i;
        c = a / i;
        break;
      }
    }
  }
  cout << b << " " << c << endl;
  return 0;
}