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.*;
import static java.lang.Math.*;
public class ProblemC {
    void run() {
        Scanner sc = new Scanner(System.in);
		while (sc.hasNextInt()) {
			int x = sc.nextInt();
			int res;
			if (x == 3) {
				res = 5;
			}
			else {
				res = 1;
				while (res * res + 1 < 2 * x) {
					res += 2;
				}
			}
			p("%d\n", res);
		}
    }

    boolean debug = false;
    void p(String f, Object...params) {
        System.out.printf(f, params);
    }
    void d(Object...params) {
        if (debug) {
            p("DEBUG: %s\n", Arrays.deepToString(params));
        }
    }
    public ProblemC(String[] args) {
        if (args.length > 0 && args[0].equals("debug")) {
            debug = true;
        }
    }
    public static void main(String[] args) {
        new ProblemC(args).run();
    }
}
