#include <bits/stdc++.h>
using namespace std;
const long long MX = 4e5 + 9;
const long double pi = acos(-1);
const long long mod = 1e9 + 7;
long long po(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = po(a, b / 2);
  x *= x % mod;
  if (b % 2) {
    x *= a % mod;
  }
  return x % mod;
}
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int dx[] = {0, 0, -1, 1, 1, -1, -1, 1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
int main() {
  int n;
  cin >> n;
  int a = 1, b = n;
  for (int i = 1; i < n; i++) {
    if (n % i == 0 && i <= n / i) {
      if (abs(i - n / i) < abs(a - b)) {
        a = i;
        b = n / i;
      }
    }
  }
  cout << a << " " << b;
  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;
const long long MX = 4e5 + 9;
const long double pi = acos(-1);
const long long mod = 1e9 + 7;
long long po(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = po(a, b / 2);
  x *= x % mod;
  if (b % 2) {
    x *= a % mod;
  }
  return x % mod;
}
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int dx[] = {0, 0, -1, 1, 1, -1, -1, 1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
int main() {
  int n;
  cin >> n;
  int a = 1, b = n;
  for (int i = 1; i < n; i++) {
    if (n % i == 0 && i <= n / i) {
      if (abs(i - n / i) < abs(a - b)) {
        a = i;
        b = n / i;
      }
    }
  }
  cout << a << " " << b;
  return 0;
}

```