#include <bits/stdc++.h>
using namespace std;
vector<long> v1;
void factors(long l);
int main() {
  long l, i;
  cin >> l;
  factors(l);
  sort(v1.begin(), v1.end());
  long prev_diff = 0;
  long n = v1.size();
  long length, breadth;
  if (n % 2 == 0) {
    cout << v1[n / 2 - 1] << " " << v1[n / 2];
  } else {
    if (v1[(n - 1) / 2] * v1[(n - 1) / 2] == l)
      cout << v1[(n - 1) / 2] << " " << v1[(n - 1) / 2];
    else
      cout << v1[(n - 1) / 2 - 1] << " " << v1[(n + 1) / 2];
  }
}
void factors(long l) {
  for (int i = 1; i <= (l / 2); i++) {
    if (l % i == 0) v1.push_back(i);
  }
  v1.push_back(l);
}

### Prompt

Please provide a CPP coded solution 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;
vector<long> v1;
void factors(long l);
int main() {
  long l, i;
  cin >> l;
  factors(l);
  sort(v1.begin(), v1.end());
  long prev_diff = 0;
  long n = v1.size();
  long length, breadth;
  if (n % 2 == 0) {
    cout << v1[n / 2 - 1] << " " << v1[n / 2];
  } else {
    if (v1[(n - 1) / 2] * v1[(n - 1) / 2] == l)
      cout << v1[(n - 1) / 2] << " " << v1[(n - 1) / 2];
    else
      cout << v1[(n - 1) / 2 - 1] << " " << v1[(n + 1) / 2];
  }
}
void factors(long l) {
  for (int i = 1; i <= (l / 2); i++) {
    if (l % i == 0) v1.push_back(i);
  }
  v1.push_back(l);
}

```