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>
#pragma warning(disable : 6031)
#pragma warning(disable : 4101)
using namespace std;
const int INF = 1e9;
const long long LINF = 1e18;
const double PI = acos(-1);
long long x;
vector<long long> factors(long long n) {
  vector<long long> fs;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      long long f = 1;
      while (n % i == 0) {
        f = f * i;
        n = n / i;
      }
      fs.push_back(f);
    }
  }
  if (n > 1) {
    fs.push_back(n);
  }
  return fs;
}
pair<long long, long long> bruteforce(vector<long long> fs,
                                      long long capacity) {
  long long best = capacity;
  pair<long long, long long> ans;
  for (int i = 0; i < (1 << (((int)(fs).size()))); i++) {
    long long cur_l = 1;
    long long cur_r = 1;
    for (int k = 0; k < ((int)(fs).size()); k++) {
      if ((i >> k) & 1) {
        cur_l *= fs[k];
      } else {
        cur_r *= fs[k];
      }
    }
    if (max(cur_l, cur_r) <= best) {
      best = max(cur_l, cur_r);
      ans = {cur_l, cur_r};
    }
  }
  return ans;
}
int main() {
  scanf("%lld", &x);
  pair<long long, long long> k = bruteforce(factors(x), x);
  printf("%lld %lld\n", k.first, k.second);
  return 0;
}
