#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c, d, n, min = 1000000, x, y;
  cin >> n;
  d = sqrt(n);
  if (n == 0) {
    cout << "0"
         << " "
         << "0" << endl;
  } else {
    for (int i = 1; i <= d; i++) {
      if (n % i == 0) {
        b = n / i;
        c = b - i;
        if (b >= i) {
          if (c <= min) {
            min = c;
            x = i;
            y = b;
          }
        } else {
          break;
        }
      }
    }
    cout << x << " " << y << endl;
  }
}

### Prompt

Please create a solution in cpp to the following problem:

A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly n pixels. 

Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels a and the number of columns of pixels b, so that:

  * there are exactly n pixels on the display; 
  * the number of rows does not exceed the number of columns, it means a ≤ b; 
  * the difference b - a is as small as possible. 

Input

The first line contains the positive integer n (1 ≤ n ≤ 106) — the number of pixels display should have.

Output

Print two integers — the number of rows and columns on the display. 

Examples

Input

8


Output

2 4


Input

64


Output

8 8


Input

5


Output

1 5


Input

999999


Output

999 1001

Note

In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.

In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.

In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c, d, n, min = 1000000, x, y;
  cin >> n;
  d = sqrt(n);
  if (n == 0) {
    cout << "0"
         << " "
         << "0" << endl;
  } else {
    for (int i = 1; i <= d; i++) {
      if (n % i == 0) {
        b = n / i;
        c = b - i;
        if (b >= i) {
          if (c <= min) {
            min = c;
            x = i;
            y = b;
          }
        } else {
          break;
        }
      }
    }
    cout << x << " " << y << endl;
  }
}

```