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 java.io.*;

public class Solution {
    FastScanner in;
    PrintWriter out;

    public void solve() throws IOException {
        int x = in.nextInt();

        if (x == 3) {
            System.out.println(5);
            return;
        }

        int i;
        for (i = 1; x > 1 + (4 * (i - 1)) * i / 2; i++) {
        }
        int n = i * 2 - 1;

        out.println(n);
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
    }

    public static void main(String[] arg) {
        new Solution().run();
    }
}