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.*;
public class Main {

    private void doit(){
        Scanner sc = new Scanner (System.in);
        while(sc.hasNext()){
            int x = sc.nextInt();
            int res = solve(x) * 2 -1;
            System.out.println(res);
        }
    }
    
    private int solve(int x){
        if(x == 1) return 1;
        if(x==2||x==4||x==5) return 2;
        int sum=5;
        for(int i=3;i < 1000;i++){
            int len = i * 2 -1;
            int inc = (len / 2 + 1) * 4 -4;
            sum += inc;
            if(sum >= x) return i;
            //System.out.println("sum= " + sum + "inc = " + inc + " len= " + len);
        }
        return -1;
    }


    public static void main(String[] args) {
        Main obj = new Main();
        obj.doit();
    }
}