#include <bits/stdc++.h>
using namespace std;
long long int n, i, l2, l4, j, k, l, ans[101];
int main() {
  cin >> n;
  for (i = 0; i <= 100; i++) ans[i] = -1;
  for (i = 101; i > 0;) {
    if ((i / 2) % 2) {
      l2 = (((i + 1) / 2) / 2 - 1) * 2;
      l4 = ((i + 1) / 2) * (((i + 1) / 2) / 2) - l2 - 1;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 1 + 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
      l2 = (((i + 1) / 2) / 2) * 2;
      l4 = ((i + 1) / 2) * (((i + 1) / 2) / 2) - l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
    } else {
      l2 = i / 2;
      l4 = ((i / 2) / 2) * l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 1 + 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
      l2 = i / 2;
      l4 = ((i / 2) / 2) * l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
    }
    i -= 2;
  }
  cout << ans[n];
  return 0;
}

### Prompt

Your challenge is to write 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;
long long int n, i, l2, l4, j, k, l, ans[101];
int main() {
  cin >> n;
  for (i = 0; i <= 100; i++) ans[i] = -1;
  for (i = 101; i > 0;) {
    if ((i / 2) % 2) {
      l2 = (((i + 1) / 2) / 2 - 1) * 2;
      l4 = ((i + 1) / 2) * (((i + 1) / 2) / 2) - l2 - 1;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 1 + 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
      l2 = (((i + 1) / 2) / 2) * 2;
      l4 = ((i + 1) / 2) * (((i + 1) / 2) / 2) - l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
    } else {
      l2 = i / 2;
      l4 = ((i / 2) / 2) * l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 1 + 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
      l2 = i / 2;
      l4 = ((i / 2) / 2) * l2;
      for (j = 0; j <= l2; j++) {
        for (k = 0; k <= l4; k++) {
          l = 2 * j + 4 * k;
          if (l > 100) break;
          ans[l] = i;
        }
      }
    }
    i -= 2;
  }
  cout << ans[n];
  return 0;
}

```