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>
using namespace std;
vector<long long> v;
int vn;
vector<long long> vl;
void put(int ind, long long val) {
  if (ind == vn) {
    vl.push_back(val);
    return;
  }
  put(ind + 1, val);
  put(ind + 1, val * v[ind]);
}
int main(void) {
  long long x;
  scanf("%lld", &x);
  if (x == 1) {
    printf("1 1");
    return 0;
  }
  for (long long i = 2; i * i <= x; i++) {
    long long k = 1;
    while (x % i == 0) {
      k *= i;
      x /= i;
    }
    if (k != 1) {
      v.push_back(k);
    }
  }
  if (x != 1) {
    v.push_back(x);
  }
  vn = v.size();
  put(0, 1);
  int vln = vl.size();
  sort(vl.begin(), vl.end());
  printf("%lld %lld", vl[vln / 2 - 1], vl[vln / 2]);
  return 0;
}