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 N = 1000005;
const long long INF = 1e18;
long long x, m, n, i, j, k;
pair<long long, long long> a[N];
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcd(long long a, long long b) { return a * b / gcd(a, b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> x;
  n = 1;
  a[1] = make_pair(1, x);
  for (i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      long long j = x / i;
      if (lcd(i, j) == x) {
        n++;
        a[n] = make_pair(i, j);
      }
    }
  }
  sort(a + 1, a + 1 + n);
  cout << a[n].first << " " << a[n].second << endl;
  return 0;
}