#include <bits/stdc++.h>
using namespace std;
int x;
inline bool check(int n, int m) {
  int now = (n * n + 1) / 2;
  if (now < m) return 0;
  int four = (n - 1) / 2, two = 2 * ((n - 1) / 4), one = 1;
  while (now > m) {
    if (now >= m + 4 && four)
      now -= 4, four--;
    else if (now >= m + 2 && two)
      now -= 2, two--;
    else if (now >= m + 1 && one)
      now -= 2, one--;
    else
      return 0;
  }
  return 1;
}
int main() {
  cin >> x;
  if (x == 1)
    puts("1");
  else if (x == 2)
    puts("3");
  else if (x == 3)
    puts("5");
  else {
    for (int i = 3; i; i += 2)
      if (check(i, x)) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 x;
inline bool check(int n, int m) {
  int now = (n * n + 1) / 2;
  if (now < m) return 0;
  int four = (n - 1) / 2, two = 2 * ((n - 1) / 4), one = 1;
  while (now > m) {
    if (now >= m + 4 && four)
      now -= 4, four--;
    else if (now >= m + 2 && two)
      now -= 2, two--;
    else if (now >= m + 1 && one)
      now -= 2, one--;
    else
      return 0;
  }
  return 1;
}
int main() {
  cin >> x;
  if (x == 1)
    puts("1");
  else if (x == 2)
    puts("3");
  else if (x == 3)
    puts("5");
  else {
    for (int i = 3; i; i += 2)
      if (check(i, x)) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}

```