#include <bits/stdc++.h>
using namespace std;
const int N = 100001;
const int MOD = 1e9 + 7;
int main(int argc, const char* argv[]) {
  int n;
  cin >> n;
  vector<int> factors;
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
    }
  }
  int x1, x2, diff = INT_MAX;
  for (int i = 0; i < factors.size(); i++) {
    for (int j = 0; j < factors.size(); j++) {
      if (factors[i] * factors[j] == n && abs(factors[i] - factors[j]) < diff) {
        diff = abs(factors[i] - factors[j]);
        x1 = factors[i];
        x2 = factors[j];
      }
    }
  }
  cout << min(x1, x2) << " " << max(x1, x2) << 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;
const int N = 100001;
const int MOD = 1e9 + 7;
int main(int argc, const char* argv[]) {
  int n;
  cin >> n;
  vector<int> factors;
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
    }
  }
  int x1, x2, diff = INT_MAX;
  for (int i = 0; i < factors.size(); i++) {
    for (int j = 0; j < factors.size(); j++) {
      if (factors[i] * factors[j] == n && abs(factors[i] - factors[j]) < diff) {
        diff = abs(factors[i] - factors[j]);
        x1 = factors[i];
        x2 = factors[j];
      }
    }
  }
  cout << min(x1, x2) << " " << max(x1, x2) << endl;
  return 0;
}

```