#include <bits/stdc++.h>
using namespace std;
void delitel(int n, int &m, int &s) {
  int k = n;
  for (int i = 1; i <= n / 2; ++i) {
    if (n % i == 0) {
      if (n / i >= i) {
        if (min(abs(n / i - i), k) == (n / i - i)) {
          m = i;
          s = n / i;
          k = n / i - i;
        }
      }
    }
  }
}
int main() {
  int n, m, s;
  cin >> n;
  delitel(n, m, s);
  if (n != 1) {
    cout << m << " " << s;
  } else {
    cout << 1 << " " << 1;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
void delitel(int n, int &m, int &s) {
  int k = n;
  for (int i = 1; i <= n / 2; ++i) {
    if (n % i == 0) {
      if (n / i >= i) {
        if (min(abs(n / i - i), k) == (n / i - i)) {
          m = i;
          s = n / i;
          k = n / i - i;
        }
      }
    }
  }
}
int main() {
  int n, m, s;
  cin >> n;
  delitel(n, m, s);
  if (n != 1) {
    cout << m << " " << s;
  } else {
    cout << 1 << " " << 1;
  }
}

```