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.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;

class Main{


	public static void main(String args[]){
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		try{
			String str;
			while((str = br.readLine()) != null){
				int n = Integer.parseInt(str);
				calc(n);
			}
		}catch(Exception e){
		}
	}

	static void calc(int n){
		switch(n){
		case 0:{
			System.out.println(0);
		}
		case 1:{
			System.out.println(1);
			break;
		}
		case 2:{
			System.out.println(2);
			break;
		}
		case 3:{
			System.out.println(5);
			break;
		}
		default:{
			if(n%2 == 0){
				int m = n/2 - 1;
				int result = 1;
				int edge = 3;
				for(int i = 0; i < m; i++){
					result += edge;
					edge *= 3;
				}
				System.out.println(result * 2);
			}else{
				int m = (n - 1) / 2;
				int result = 1;
				int edge = 4;
				for(int i = 0; i < m; i++){
					result += edge;
					edge *= 3;
				}
				System.out.println(result);
			}
		}
		}
	}
}