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;
vector<long long> factors;
void trial(long long n) {
  int count;
  long long ini = n;
  for (long long d = 2; d * d <= n; d++) {
    if (n % d == 0) {
      ini = n;
      while (n % d == 0) {
        n /= d;
      }
      long long s = ini / n;
      factors.push_back(s);
    }
  }
  if (n > 1) {
    factors.push_back(n);
  }
}
int main() {
  long long n;
  cin >> n;
  trial(n);
  int siz = factors.size();
  long long a = 1;
  long long b = 1;
  long long ra = 1, rb = 1;
  long long mini = 1000000000000;
  for (int i = 1; i < (1 << siz); i++) {
    a = 1;
    b = 1;
    for (int j = 1; j <= siz; j++) {
      if (i & (1 << (j - 1)))
        a *= factors[siz - j];
      else
        b *= factors[siz - j];
    }
    if (mini > max(a, b)) {
      ra = a;
      rb = b;
      mini = max(a, b);
    }
  }
  cout << ra << ' ' << rb << endl;
}