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.*;
    import java.math.*;
    
    public class C {
        public static Scanner scan = new Scanner(System.in);
        public static boolean bg = true;
    
        public static void main(String[] args) throws Exception {
            TreeMap<Integer,Integer> odd = new TreeMap();
            int cur = 1;
            for (int i=1;i<=100;i+=2){
                odd.put(cur, i);
                cur+=(i+1)*2;
            }
            
            TreeMap<Integer,Integer> even = new TreeMap();
            for (int i=2;i<=100;i+=2){
                int k1 = i/4;
                int add1 = k1*k1*4;
                
                
                
                int k2 = (i-1)/4;
                int add2 = k2*k2*4;
                even.put(add1+add2, i);
            }
            //if (bg) System.out.println(odd);
            //if (bg) System.out.println(even);
            int k1 = Integer.parseInt(scan.next());
            System.out.println(ans(k1));
            
        }
        public static int ans (int n1){
            if (n1==1) return 1;
            if (n1%4==0){
                int k1 = n1/4;
                for (int i=0;i<=1000;i++){
                    k1 -= i;
                    if (k1<=0) return i*2+1;
                }
            }
            else if (n1%4==1){
                
                return ans(n1-1);
            }
            else if (n1%4==2){
                return (ans(n1+2));
            }
            else if (n1%4==3){
                if (n1==3) return 5;
                else {
                    return (ans(n1+2));
                }
            }
            return -1;
        }
    }
    
    
