We have a cord whose length is a positive integer. We will perform the following condition until the length of the cord becomes at most 2:

* Operation: Cut the rope at two positions to obtain three cords, each with a length of a positive integer. Among these, discard one with the longest length and one with the shortest length, and keep the remaining one.



Let f(N) be the maximum possible number of times to perform this operation, starting with a cord with the length N.

You are given a positive integer X. Find the maximum integer N such that f(N)=X.

Constraints

* 1 \leq X \leq 40

Input

The input is given from Standard Input in the following format:


X


Output

Print the value of the maximum integer N such that f(N)=X.

Example

Input

2


Output

14
#include<bits/stdc++.h>
#include <stdlib.h>
#include <stdio.h>
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define int long long
#define vec(a,n) vector<int> (a)((n));
#define Vec(a,n) vector<string> (a)((n));
#define twovec(a,n,m) vector<vector<int>> a(n,vector<int>(m,0));
#define Twovec(a,n,m) vector<vector<double>> a(n,vector<double>(m,0));
#define P pair<int,int>
#define All(a) (a).begin(),(a).end()
#define Sort(a) sort(All(a));
#define Reverse(a) reverse(All(a));
#define PQ(n) priority_queue<P,vector<P>,greater<P>> (n)
#define pq(n) priority_queue<int> (n)
#define print(a) cout << (a) << endl;
#define printD(a) cout << setprecision(10) << ans << endl;
using namespace std;
int max_int = 1000000007;
void Debug(auto a);
int nibul(auto a,auto b);
int nibuu(auto a,auto b);
void input(vector<int>& a,int n);

int n,m;
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};


signed main(){
    cin >> n;
    int ans = n*2+2;
    print(ans*2+2)
    return 0;
}

void Debug(auto a){
    cout << "{ ";
    for(auto b: a){
        cout << b << " ";
    }
    cout << "}" << endl;
}
int nibul(auto a,auto b){
    int x = lower_bound(All(a),b) - a.begin();
    //key以上の初めてのitr
    return x;
}
int nibuu(auto a,auto b){
    int x = upper_bound(All(a),b) - a.begin();
    //key以下の最後のitr
    return x-1;
}

void input(vector<int>& a,int n){
    REP(i,n){
        cin >> a[i];
    }
}
