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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios_base::sync_with_stdio(false);
  long long x;
  cin >> x;
  vector<long long> v;
  long long u = sqrt(x);
  for (long long i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      v.push_back(i);
      long long y = x / i;
      if (y > u) v.push_back(y);
    }
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    for (int j = i; j >= 0; j--) {
      if (lcm(v[i], v[j]) == x) {
        cout << v[i] << " " << v[j] << endl;
        return 0;
      }
    }
  }
  return 0;
}