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.

<image>
---
Figure 1: These three examples of isomers of heptane have the same number of carbons but different structures.
---



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.

<image>
---
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.
---



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
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    vector<int> ans(31, 0);

    ans[0] = 0;
    for(int i = 1; i <= 30; ++i) {
        ans[i] = i;
        for(int j = 1; j < i-1; ++j) {
            int level = min(j, i-j-1);
            int x = 1;
            for(int k = 0; k < level; ++k) {
                ans[i] += 2*x;
                x *= 3;
            }
        }
    }

    int N;
    while(cin >> N) {
        cout << ans[N] << endl;
    }
    return 0;
}