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>
long long max(long long a, long long b) { return (a < b) ? b : a; }
long long min(long long a, long long b) { return (a > b) ? b : a; }
using namespace std;
long long x[300], y[300];
long long ab(long long r) { return (r < 0) ? -r : r; }
int main() {
  long long x, ans = LLONG_MAX / 2, bb, aa;
  cin >> x;
  map<long long, long long> pr;
  while (x > 1) {
    long long br = 1;
    for (long long i = 2; i * i <= x; i++) {
      if (x % i == 0) {
        pr[i]++;
        x /= i;
        br = 0;
        break;
      }
    }
    if (br) {
      pr[x]++;
      x = 1;
    }
  }
  for (long long i = 0; i < 1 << pr.size(); i++) {
    long long a = 1, b = 1;
    long long te = i;
    for (auto&& j : pr) {
      if (te % 2 == 1) {
        for (long long v = 0; v < j.second; v++) a *= j.first;
      } else {
        for (long long v = 0; v < j.second; v++) b *= j.first;
      }
      te /= 2;
    }
    if (ans > max(a, b)) ans = max(a, b), aa = a, bb = b;
  }
  cout << aa << " " << bb << endl;
}