#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  switch (n) {
    case 1:
      cout << 1;
      break;
    case 2:
      cout << 3;
      break;
    case 3:
      cout << 5;
      break;
    case 4:
    case 5:
      cout << 3;
      break;
    case 6:
    case 7:
      cout << 5;
      break;
    default:
      int n4 = n / 4;
      if (n4 % 2)
        n4 = n4 * 2 - 1;
      else
        n4 = n4 * 2;
      int row = sqrt(n4);
      if (row * row < n4) ++row;
      int a = (row - 1) * (row - 1);
      int m = a / 2;
      if (m * 2 < a) m++;
      int l = n - m * 4;
      if (l <= 1 + ((row - 1) / 2) * 4) row--;
      cout << row * 2 + 1;
  }
}

### Prompt

Please formulate a cpp solution 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 main() {
  int n;
  cin >> n;
  switch (n) {
    case 1:
      cout << 1;
      break;
    case 2:
      cout << 3;
      break;
    case 3:
      cout << 5;
      break;
    case 4:
    case 5:
      cout << 3;
      break;
    case 6:
    case 7:
      cout << 5;
      break;
    default:
      int n4 = n / 4;
      if (n4 % 2)
        n4 = n4 * 2 - 1;
      else
        n4 = n4 * 2;
      int row = sqrt(n4);
      if (row * row < n4) ++row;
      int a = (row - 1) * (row - 1);
      int m = a / 2;
      if (m * 2 < a) m++;
      int l = n - m * 4;
      if (l <= 1 + ((row - 1) / 2) * 4) row--;
      cout << row * 2 + 1;
  }
}

```