#include <bits/stdc++.h>
using namespace std;
bool p[1000010];
int main() {
  for (int i = 2; i <= 50000; i++)
    for (int j = 2; j * i <= 1000000; j++) {
      p[i * j] = 1;
    }
  int n;
  scanf("%d", &n);
  int q = sqrt(n);
  if (p[n] == 0)
    printf("1 %d\n", n);
  else if (q * q == n)
    printf("%d %d\n", q, q);
  else {
    int s, q2 = q;
    bool flag = 0;
    while (1) {
      s = q2 + 1;
      while (1) {
        if (q2 * s == n) {
          flag = 1;
          break;
        } else if (q2 * s > n)
          break;
        else
          s++;
      }
      if (flag)
        break;
      else
        q2--;
    }
    printf("%d %d\n", q2, s);
  }
  return 0;
}

### 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;
bool p[1000010];
int main() {
  for (int i = 2; i <= 50000; i++)
    for (int j = 2; j * i <= 1000000; j++) {
      p[i * j] = 1;
    }
  int n;
  scanf("%d", &n);
  int q = sqrt(n);
  if (p[n] == 0)
    printf("1 %d\n", n);
  else if (q * q == n)
    printf("%d %d\n", q, q);
  else {
    int s, q2 = q;
    bool flag = 0;
    while (1) {
      s = q2 + 1;
      while (1) {
        if (q2 * s == n) {
          flag = 1;
          break;
        } else if (q2 * s > n)
          break;
        else
          s++;
      }
      if (flag)
        break;
      else
        q2--;
    }
    printf("%d %d\n", q2, s);
  }
  return 0;
}

```