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.io.*;
import java.util.*;


public class C {
    public static void main(String[] args) throws Exception {
        new C().solve();
    }
    void solve() throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int x = Integer.parseInt(in.readLine());
        if (x == 1) {
            System.out.println(1);
            return;
        } else if (x == 3) {
            System.out.println(5);
            return;
        } else {
            for (int n = 3;; n++) {
                if (n % 2 == 1) {
                    // odd
                    if (x <= n * n / 2 + 1) {
                        System.out.println(n);
                        return;
                    }
                } else {
                    // even
                    int k = n / 2 - 1;
                    int upper = k % 2 == 0 ? k * k / 2 : k * k / 2 + 1;
                    if (x % 4 == 0 && x <= 4 * upper) {
                        System.out.println(n);
                        return;
                    }
                }
            }

        }
    }
}
