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
#include <bits/stdc++.h>
using namespace std;
vector<long long> v, vc;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  long long sm = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      long long c = 0, d = 1;
      while (n % i == 0) {
        n /= i;
        c++;
        d *= i;
      }
      v.push_back(d);
    }
  }
  if (n > 1) {
    v.push_back(n);
  }
  long long l = v.size();
  long long m = (1LL << l);
  for (long long i = 0; i < m; i++) {
    long long p = 1;
    for (long long j = 0; j < l; j++) {
      if (i & (1LL << j)) {
        p *= v[j];
      }
    }
    vc.push_back(p);
  }
  long long x1, x2;
  long long mx = 1e18;
  for (long long u : vc) {
    long long p = sm / u;
    long long q = max(u, p);
    if (mx > q) {
      mx = q;
      x1 = u, x2 = p;
    }
  }
  if (x1 > x2) swap(x1, x2);
  cout << x1 << " " << x2 << "\n";
  return 0;
}
