#include <bits/stdc++.h>
int f(int n) {
  if (n == 0) return 0;
  if (n & 1)
    return (n * n + 1) / 2;
  else {
    int k = n / 2 - 1;
    if (k & 1)
      return 2 * (k * k + 1);
    else
      return 2 * (k * k);
  }
}
int main() {
  int x, i;
  while (~scanf("%d", &x)) {
    if (x == 1) {
      printf("%d\n", 1);
      continue;
    }
    if (x == 5) {
      printf("%d\n", 3);
      continue;
    }
    for (i = 0; i <= 10007; i++)
      if (x <= f(i)) {
        if (i & 1)
          if ((f(i) - x - 1) % 2 == 0 || i > 4 && (f(i) - x) % 2 == 0) {
            printf("%d\n", i);
            break;
          } else
            ;
        else if ((f(i) - x) % 4 == 0) {
          printf("%d\n", i);
          break;
        }
      }
  }
}

### Prompt

Generate 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>
int f(int n) {
  if (n == 0) return 0;
  if (n & 1)
    return (n * n + 1) / 2;
  else {
    int k = n / 2 - 1;
    if (k & 1)
      return 2 * (k * k + 1);
    else
      return 2 * (k * k);
  }
}
int main() {
  int x, i;
  while (~scanf("%d", &x)) {
    if (x == 1) {
      printf("%d\n", 1);
      continue;
    }
    if (x == 5) {
      printf("%d\n", 3);
      continue;
    }
    for (i = 0; i <= 10007; i++)
      if (x <= f(i)) {
        if (i & 1)
          if ((f(i) - x - 1) % 2 == 0 || i > 4 && (f(i) - x) % 2 == 0) {
            printf("%d\n", i);
            break;
          } else
            ;
        else if ((f(i) - x) % 4 == 0) {
          printf("%d\n", i);
          break;
        }
      }
  }
}

```