#include <bits/stdc++.h>
using namespace std;
int x;
int f(int a, int b) { return (a * b + 1) / 2; }
bool ok(int n) {
  if (x == 3 && n == 3) return false;
  if (n % 2 == 0 && x % 4) return false;
  if (n % 2 == 0) {
    if (x / 4 <= f(n / 2 - 1, n / 2 - 1)) return true;
    return false;
  }
  int z = f(n, n);
  if (x > z) return false;
  return true;
}
int main() {
  cin >> x;
  for (int i = (int)(1), _n = (int)(100000); i < _n; i++)
    if (ok(i)) {
      cout << i << endl;
      break;
    }
  {
    int _;
    cin >> _;
  }
}

### Prompt

Please create a solution in CPP to the following problem:

Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as Ai, j.

Let's call matrix A clear if no two cells containing ones have a common side.

Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: Ai, j = An - i + 1, j and Ai, j = Ai, n - j + 1.

Let's define the sharpness of matrix A as the number of ones in it.

Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.

Input

The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.

Output

Print a single number — the sought value of n.

Examples

Input

4


Output

3


Input

9


Output

5

Note

The figure below shows the matrices that correspond to the samples:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int x;
int f(int a, int b) { return (a * b + 1) / 2; }
bool ok(int n) {
  if (x == 3 && n == 3) return false;
  if (n % 2 == 0 && x % 4) return false;
  if (n % 2 == 0) {
    if (x / 4 <= f(n / 2 - 1, n / 2 - 1)) return true;
    return false;
  }
  int z = f(n, n);
  if (x > z) return false;
  return true;
}
int main() {
  cin >> x;
  for (int i = (int)(1), _n = (int)(100000); i < _n; i++)
    if (ok(i)) {
      cout << i << endl;
      break;
    }
  {
    int _;
    cin >> _;
  }
}

```