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 int MAXM = 1e2 + 5;
inline void chk_min(long long &a, long long b) {
  if (a > b) a = b;
}
inline void chk_max(long long &a, long long b) {
  if (a < b) a = b;
}
int p[MAXM], m = 0;
long long q[MAXM];
int main(void) {
  long long n;
  scanf("%lld", &n);
  long long sn = sqrt(n), t = n;
  for (int i = 2; i <= sn && t > 1; ++i)
    if (t % i == 0) {
      p[m] = i;
      q[m] = 1;
      while (t % i == 0) t /= i, q[m] *= (long long)i;
      ++m;
    }
  if (t > 1) q[m++] = t;
  long long ans = n;
  int all = (1 << m) - 1;
  for (int s = 0; s <= all; ++s) {
    long long tmp = 1;
    for (int i = 0; i < m; ++i)
      if ((((s) >> (i)) & 1)) tmp *= q[i];
    chk_min(ans, max(tmp, n / tmp));
  }
  printf("%lld %lld", ans, n / ans);
  return 0;
}
