#include <bits/stdc++.h>
using namespace std;
int x, n;
int main() {
  scanf("%d", &x);
  if (x == 3) {
    printf("%d\n", 5);
    return 0;
  };
  if (x & 1) {
    if ((x - 1) % 4 == 0) {
      for (n = 1;; n++)
        if ((n & 1) && n * n / 2 + 1 >= x) {
          printf("%d\n", n);
          return 0;
        };
    } else {
      for (n = 1;; n++)
        if ((n & 1) && n * n / 2 - 1 >= x) {
          printf("%d\n", n);
          return 0;
        };
    }
  } else {
    if (x % 4 == 0) {
      for (n = 1;; n++)
        if (n & 1) {
          if (n * n / 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
        } else {
          if ((n - 2) / 2 * (n - 2) / 2 / 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
        }
    } else {
      for (n = 1;; n++)
        if (n & 1)
          if (n * n / 2 - 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
    }
  }
}

### Prompt

Create a solution in Cpp for 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, n;
int main() {
  scanf("%d", &x);
  if (x == 3) {
    printf("%d\n", 5);
    return 0;
  };
  if (x & 1) {
    if ((x - 1) % 4 == 0) {
      for (n = 1;; n++)
        if ((n & 1) && n * n / 2 + 1 >= x) {
          printf("%d\n", n);
          return 0;
        };
    } else {
      for (n = 1;; n++)
        if ((n & 1) && n * n / 2 - 1 >= x) {
          printf("%d\n", n);
          return 0;
        };
    }
  } else {
    if (x % 4 == 0) {
      for (n = 1;; n++)
        if (n & 1) {
          if (n * n / 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
        } else {
          if ((n - 2) / 2 * (n - 2) / 2 / 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
        }
    } else {
      for (n = 1;; n++)
        if (n & 1)
          if (n * n / 2 - 2 >= x) {
            printf("%d\n", n);
            return 0;
          };
    }
  }
}

```