#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m1 = 0, m2 = 0, diferencia = 0;
  cin >> n;
  diferencia = n;
  if ((n % 2 == 0) and n != 0) {
    for (int iA = 0; iA < n; iA += 2) {
      if (iA == 0) {
        m1 = 1;
        m2 = n;
      } else {
        if (iA * (n / iA) == n) {
          if ((iA > (n / iA)) and (abs(iA - n / iA) < diferencia)) {
            m2 = iA;
            m1 = n / iA;
            diferencia = abs(iA - n / iA);
          } else if ((iA <= (n / iA)) and (abs(iA - n / iA) < diferencia)) {
            m1 = iA;
            m2 = n / iA;
            diferencia = abs(iA - n / iA);
          }
        }
      }
    }
  } else if ((n % 2 != 0) and n != 0 and n != 1) {
    for (int iA = 1; iA < n; iA += 2) {
      if (iA * (n / iA) == n) {
        if ((iA > (n / iA)) and (abs(iA - n / iA) < diferencia)) {
          m2 = iA;
          m1 = n / iA;
          diferencia = abs(iA - n / iA);
        } else if ((iA <= (n / iA)) and (abs(iA - n / iA) < diferencia)) {
          m1 = iA;
          m2 = n / iA;
          diferencia = abs(iA - n / iA);
        }
      }
    }
  } else if (n == 1) {
    m1 = 1;
    m2 = 1;
  }
  cout << m1 << " " << m2 << endl;
  return 0;
}

### Prompt

Generate a CPP solution 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 n, m1 = 0, m2 = 0, diferencia = 0;
  cin >> n;
  diferencia = n;
  if ((n % 2 == 0) and n != 0) {
    for (int iA = 0; iA < n; iA += 2) {
      if (iA == 0) {
        m1 = 1;
        m2 = n;
      } else {
        if (iA * (n / iA) == n) {
          if ((iA > (n / iA)) and (abs(iA - n / iA) < diferencia)) {
            m2 = iA;
            m1 = n / iA;
            diferencia = abs(iA - n / iA);
          } else if ((iA <= (n / iA)) and (abs(iA - n / iA) < diferencia)) {
            m1 = iA;
            m2 = n / iA;
            diferencia = abs(iA - n / iA);
          }
        }
      }
    }
  } else if ((n % 2 != 0) and n != 0 and n != 1) {
    for (int iA = 1; iA < n; iA += 2) {
      if (iA * (n / iA) == n) {
        if ((iA > (n / iA)) and (abs(iA - n / iA) < diferencia)) {
          m2 = iA;
          m1 = n / iA;
          diferencia = abs(iA - n / iA);
        } else if ((iA <= (n / iA)) and (abs(iA - n / iA) < diferencia)) {
          m1 = iA;
          m2 = n / iA;
          diferencia = abs(iA - n / iA);
        }
      }
    }
  } else if (n == 1) {
    m1 = 1;
    m2 = 1;
  }
  cout << m1 << " " << m2 << endl;
  return 0;
}

```