#include <bits/stdc++.h>
using namespace std;
const int maxx = 1e6 + 19;
long long n, m, a[maxx], dp[maxx], b[maxx], ans, jvb, sum, c, v0, v1, l, x, y;
string s, t;
bool flag = true;
vector<long long> v;
void Input() { cin >> n; }
void Solve() {
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) v.push_back(i);
  }
  for (int i = 0; i < v.size(); i++) {
    for (int j = i; j < v.size(); j++) {
      if (v[i] * v[j] == n) {
        x = v[i];
        y = v[j];
      }
    }
  }
  cout << x << " " << y << endl;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  Input(), Solve();
  return 0;
}

### 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;
const int maxx = 1e6 + 19;
long long n, m, a[maxx], dp[maxx], b[maxx], ans, jvb, sum, c, v0, v1, l, x, y;
string s, t;
bool flag = true;
vector<long long> v;
void Input() { cin >> n; }
void Solve() {
  for (int i = 1; i <= n; i++) {
    if (n % i == 0) v.push_back(i);
  }
  for (int i = 0; i < v.size(); i++) {
    for (int j = i; j < v.size(); j++) {
      if (v[i] * v[j] == n) {
        x = v[i];
        y = v[j];
      }
    }
  }
  cout << x << " " << y << endl;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  Input(), Solve();
  return 0;
}

```