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 mod = 1000000007;
long long gcd(long long a, long long b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0),
      cin.exceptions(cin.failbit);
  long long x, a, b;
  cin >> x;
  if (x == 1) {
    cout << 1 << " " << 1 << '\n';
    return 0;
  }
  vector<long long> v;
  for (int i = 1; i <= sqrt(x) + 1; i++) {
    if (x % i == 0) {
      v.push_back(i);
      v.push_back(x / i);
    }
  }
  long long ans = 10000000000000000;
  for (int i = 0; i < v.size() - 1; i++) {
    long long x1 = v[i];
    long long x2 = v[i + 1];
    if (((x1 * 1ll * x2) / gcd(x1, x2)) == x) {
      long long mx = max(v[i], v[i + 1]);
      if (mx < ans) {
        a = v[i];
        b = v[i + 1];
        ans = mx;
      }
    }
  }
  cout << a << " " << b << '\n';
  return 0;
}