#include <bits/stdc++.h>
using namespace std;
bool Check(int four, int two, int one, int tmp) {
  while (tmp >= 4 && four) tmp -= 4, --four;
  while (tmp >= 2 && two) tmp -= 2, --two;
  while (tmp >= 1 && one) tmp -= 1, --one;
  return tmp == 0;
}
int main() {
  int x;
  cin >> x;
  if (x == 1) return puts("1"), 0;
  for (register int i = 3; 666; i += 2) {
    int tmp = (i + 1) / 2;
    if (tmp & 1) {
      if (Check((tmp - 1) * (tmp - 1) / 2, (tmp - 1) * 2 / 2, 1, x))
        return cout << i << endl, 0;
    } else {
      if (Check((tmp - 1) * (tmp - 1) / 2 + 1, (tmp - 1) / 2 * 2, 1, x) ||
          Check((tmp - 1) * (tmp - 1) / 2, ((tmp - 1) / 2 + 1) * 2, 0, x))
        return cout << i << endl, 0;
    }
  }
  return 0;
}

### 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;
bool Check(int four, int two, int one, int tmp) {
  while (tmp >= 4 && four) tmp -= 4, --four;
  while (tmp >= 2 && two) tmp -= 2, --two;
  while (tmp >= 1 && one) tmp -= 1, --one;
  return tmp == 0;
}
int main() {
  int x;
  cin >> x;
  if (x == 1) return puts("1"), 0;
  for (register int i = 3; 666; i += 2) {
    int tmp = (i + 1) / 2;
    if (tmp & 1) {
      if (Check((tmp - 1) * (tmp - 1) / 2, (tmp - 1) * 2 / 2, 1, x))
        return cout << i << endl, 0;
    } else {
      if (Check((tmp - 1) * (tmp - 1) / 2 + 1, (tmp - 1) / 2 * 2, 1, x) ||
          Check((tmp - 1) * (tmp - 1) / 2, ((tmp - 1) / 2 + 1) * 2, 0, x))
        return cout << i << endl, 0;
    }
  }
  return 0;
}

```