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;
const int maxn = 2e5 + 10;
void prework() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
struct pair {
  long long x, y;
} arr[maxn];
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  prework();
  long long x;
  cin >> x;
  long long len = 0;
  for (long long i = 1; i <= sqrt(x) + 1; i++) {
    if (x % i == 0) {
      arr[len].x = i;
      arr[len++].y = x / i;
    }
  }
  long long x1 = arr[0].x, x2 = arr[0].y;
  for (long long i = 1; i < len; i++) {
    if (arr[i].x == arr[i].y) continue;
    if (lcm(arr[i].x, arr[i].y) != x) continue;
    long long cur1 = max(arr[i].x, arr[i].y);
    long long cur2 = max(x1, x2);
    if (cur2 > cur1) {
      x1 = arr[i].x;
      x2 = arr[i].y;
    }
  }
  cout << x1 << " " << x2 << endl;
  return 0;
}