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 int INF = 1e9 + 10;
const long long int LINF = 1LL * INF * INF;
const int MAXN = 2e5 + 10;
const int MAXM = 5e3 + 10;
priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;
long long int mygcd(long long int a, long long int b) {
  return a ? mygcd(b % a, a) : b;
}
int main() {
  long long int n, m, k, a, b, x, y, q;
  int sum = 0;
  int cnt = 0;
  int mx = 0;
  int mn = INF;
  int cur = 0, idx = -1;
  int tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  long long int res = n;
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    a = mygcd(i, n / i);
    if (a == 1) {
      res = min(n / i, res);
    }
  }
  cout << n / res << " " << res << "\n";
  return 0;
}