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>
import java.util.Scanner;

public class Main
{
    public static void main( String[] args )
    {
        Scanner in = new Scanner( System.in );
        int x = in.nextInt();
        if ( x > 1 && x < 6 && x != 3 ) System.out.println( 3 );
        else if ( x == 1 ) System.out.println( 1 );
        else
        {
        int current_max_of_1 = 6;
        int optimal_length = 5;
        for ( int i = 5; current_max_of_1 < x ; i += 2 )
        {
            current_max_of_1 = ( i / 2 + 1 ) * ( i / 2 + 1 ) + ( i / 2 ) * ( i / 2 );
            optimal_length = i;
        }
        System.out.println( optimal_length );
        }
    }
}
