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>
//package Round_127;

import java.util.Scanner;

public class c_beta {
	
	static long sum1(int n){
		long sum = n;
		n-=2;
		while (n > 0){
			sum += n*2;
			n-=2;
		}
		return sum;
	}
	
	static long sum2(int n){
		n--;
		long sum = 0;
		while (n > 0){
			sum += n*2;
			n-=2;
		}
		return sum;
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int x = in.nextInt();
		int n = 1;
		if (x == 2){
			System.out.println(3);
			return;
		}
		if (x == 3){
			System.out.println(5);
			return;
		}
		int i = 1;
		for (;;i+=2){
			int t = (i*i + 1) / 2;
			if (t >= x){
				System.out.println(i);
				return;
			}
		}
		
		//n = (x / 4)*2  + (x % 4 == 1 || x % 4 == 2 ? 1 : x % 4 != 0 ? 5 : 1);
		//System.out.println(n);
	}

}
