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;
const long long mod = 1e9 + 7;
const long long modl = 998244353;
const double pie = 3.1415926535;
unsigned long long power(unsigned long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else {
    if (y % 2 == 0)
      return power(x * x, y / 2);
    else
      return x * power(x * x, (y - 1) / 2);
  }
}
long long mod_power(long long x, long long y, long long m) {
  long long r = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) r = (r * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return r;
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  while (TESTS--) {
    long long x;
    scanf("%lld", &x);
    long long a = 0;
    for (long long i = 1; i * i <= x; i++) {
      if (x % i == 0 && gcd(i, x / i) == 1) {
        a = i;
      }
    }
    printf("%lld %lld", a, x / a);
  }
}
