#include <bits/stdc++.h>
int most[4][20];
int main() {
  memset(most, 0, sizeof(most));
  for (int i = 0; i < 10; ++i) {
    most[0][i] = i * i + (i + 1) * (i + 1) - 1;
    most[1][i] = i * i + (i + 1) * (i + 1);
    most[2][i] = i * i + (i + 1) * (i + 1) - 3;
    most[3][i] = i * i + (i + 1) * (i + 1) - 2;
  }
  most[3][1] = 0;
  int x;
  scanf("%d", &x);
  for (int i = 0; i < 10; ++i) {
    if (most[x % 4][i] >= x) {
      printf("%d\n", 2 * i + 1);
      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>
int most[4][20];
int main() {
  memset(most, 0, sizeof(most));
  for (int i = 0; i < 10; ++i) {
    most[0][i] = i * i + (i + 1) * (i + 1) - 1;
    most[1][i] = i * i + (i + 1) * (i + 1);
    most[2][i] = i * i + (i + 1) * (i + 1) - 3;
    most[3][i] = i * i + (i + 1) * (i + 1) - 2;
  }
  most[3][1] = 0;
  int x;
  scanf("%d", &x);
  for (int i = 0; i < 10; ++i) {
    if (most[x % 4][i] >= x) {
      printf("%d\n", 2 * i + 1);
      return 0;
    }
  }
}

```