An bydrocarbon is an organic compound which contains only carbons and hydrogens. An isomer is a compound that has the same number of carbons but different structures. Heptane, for example, is a hydrocarbon with 7 carbons. It has nine isomers. The structural formula of three are shown in Figure 1. Carbons are represented by the letter C, and bonds between carbons are represented by a straight line. In all figures, hydrogens are not represented for simplicity. Each carbon can be connected to a maximum of 4 carbons.

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Figure 1: These three examples of isomers of heptane have the same number of carbons but different structures.
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Let define a chain in an isomer as a sequence of connected carbons without branches. An isomer can have many chains of the same length. Figure 2 shows the longest chain of carbons for each of the represented isomer. Note that there can be many instances of longest chain in an isomer.

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Figure 2: The figures shows one instance of longest chain of carbons in each isomer. The first and the second isomers show longest chains of 5 carbons. The longest chain in the third isomer has 4 carbons.
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Your task is to identify the number of carbons of the largest possible carbon compound whose longest carbon chain has n (1 ≤ n ≤ 30) carbons.



Input

Each input contains a list of number of carbons, i.e. the length of a carbon chain.

The number of input lines is less than or equal to 30.

Output

For each number n in input, identify the number of carbons of the largest possible carbon compound whose longest carbon chain has n carbons.

Example

Input

1
4


Output

1
8
import java.io.*;
public class Main{
        public static void main(String[] args) throws IOException {
                BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
                String is; 
                while((is = in.readLine()) != null){
                        int n = Integer.parseInt(is);
                        if(n == 0){
                                System.out.println(0);
                                continue;
                        } else if(n == 1){
                                System.out.println(1);
                                continue;
                        } else if(n == 2){
                                System.out.println(2);
                                continue;
                        }
                        int c = 2;
                        int d = 0;
                        for(int i = 2;i < n; ++i){
                                if(i % 2 == 0){
                                        d = d * 3 + 1;
                                }
                                c += d * 2 + 1;
                        }
                        System.out.println(c);
                }   
        }   
}