#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, mi = 10000005, a, b;
  vector<pair<long long, long long> > m;
  cin >> n;
  double data = sqrt(n);
  if (data == (long long)data)
    a = b = (long long)data;
  else {
    for (long long i = 1; i <= n; i++) {
      data = (double)n / i;
      if (data == (long long)data) m.push_back(make_pair(i, data));
    }
    for (auto i : m)
      if (fabs(i.first - i.second) < mi) {
        mi = fabs(i.first - i.second);
        a = i.first, b = i.second;
      }
    if (a > b) swap(a, b);
  }
  cout << a << ' ' << b << '\n';
}

### Prompt

Create a solution in CPP for 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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, mi = 10000005, a, b;
  vector<pair<long long, long long> > m;
  cin >> n;
  double data = sqrt(n);
  if (data == (long long)data)
    a = b = (long long)data;
  else {
    for (long long i = 1; i <= n; i++) {
      data = (double)n / i;
      if (data == (long long)data) m.push_back(make_pair(i, data));
    }
    for (auto i : m)
      if (fabs(i.first - i.second) < mi) {
        mi = fabs(i.first - i.second);
        a = i.first, b = i.second;
      }
    if (a > b) swap(a, b);
  }
  cout << a << ' ' << b << '\n';
}

```