#include <bits/stdc++.h>
using namespace std;
const long long inf = LONG_MAX;
const long long arr = 1000000;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
int main(void) {
  long long mna = 0;
  long long mnb = 0;
  long long mn = LLONG_MAX;
  long long n;
  cin >> n;
  for (int i = 1; i <= arr; i++) {
    long long j = n / i;
    if (i * j == n) {
      if (abs(i - j) < mn) {
        mna = i;
        mnb = j;
        mn = abs(i - j);
      }
    }
  }
  cout << mna << " " << mnb << '\n';
  return 0;
}

### 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;
const long long inf = LONG_MAX;
const long long arr = 1000000;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
int main(void) {
  long long mna = 0;
  long long mnb = 0;
  long long mn = LLONG_MAX;
  long long n;
  cin >> n;
  for (int i = 1; i <= arr; i++) {
    long long j = n / i;
    if (i * j == n) {
      if (abs(i - j) < mn) {
        mna = i;
        mnb = j;
        mn = abs(i - j);
      }
    }
  }
  cout << mna << " " << mnb << '\n';
  return 0;
}

```