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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Solver {

    StringTokenizer st;
    BufferedReader in;
    PrintWriter out;

    public static void main(String[] args) throws NumberFormatException, IOException {
        Solver solver = new Solver();
        solver.open();
        long time = System.currentTimeMillis();
        solver.solve();
        if (!"true".equals(System.getProperty("ONLINE_JUDGE"))) {
            System.out.println("Spent time: " + (System.currentTimeMillis() - time));
        }
        solver.close();
    }

    public void open() throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
    }

    public String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws NumberFormatException, IOException {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws NumberFormatException, IOException {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws NumberFormatException, IOException {
        return Double.parseDouble(nextToken());
    }
    
    int getS(int n){
        return n*n/2+1;
    }

    public void solve() throws NumberFormatException, IOException {
        int x = nextInt();
        if (x==1){
            out.println(1);
            return;
        }else if (x==2){
            out.println(3);
            return;
        }else if (x==3){
            out.println(5);
            return;
        }
        int n = 3;
        while (getS(n)<x) n+=2;
        out.println(n);
    }

    public void close() {
        out.flush();
        out.close();
    }

}