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
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <cstring>
#include <sstream>
#include <cassert>
using namespace std;
static const double EPS = 1e-5;
typedef long long ll;
typedef pair<int,int> PI;
typedef vector<int> vi;
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
#define mp make_pair
#define pb push_back
#define f first
#define s second




int dfs(int k){
  int ret=0;
  if(k==0)return 0;
  if(k==1)ret+=1;
  else{
    ret+=1+dfs(k-1)*3;
  }
  return ret;
}

int main(){
  int n;
  while(cin>>n){
    int h1=n/2;
    int h2=n-h1;
    cout<<dfs(h1)+dfs(h2)<<endl;
  }
}