You are given a tetrahedron. Let's mark its vertices with letters A, B, C and D correspondingly.

<image>

An ant is standing in the vertex D of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.

You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex D to itself in exactly n steps. In other words, you are asked to find out the number of different cyclic paths with the length of n from vertex D to itself. As the number can be quite large, you should print it modulo 1000000007 (109 + 7). 

Input

The first line contains the only integer n (1 ≤ n ≤ 107) — the required length of the cyclic path.

Output

Print the only integer — the required number of ways modulo 1000000007 (109 + 7).

Examples

Input

2


Output

3


Input

4


Output

21

Note

The required paths in the first sample are: 

  * D - A - D
  * D - B - D
  * D - C - D
#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
const int d4i[4] = {-1, 0, 1, 0}, d4j[4] = {0, 1, 0, -1};
const int d8i[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          d8j[8] = {0, 1, 1, 1, 0, -1, -1, -1};
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
long long modpower(long long base, long long power, long long mod) {
  if (base == 0 || base == 1) return base;
  long long res = 1;
  while (power) {
    if (power % 2) {
      res = (res * base) % mod;
      power--;
    }
    base = (base * base) % mod;
    power /= 2;
  }
  return (res + mod) % mod;
}
long long const INF = 1LL << 60LL;
double const pi = 3.14159265358979323846264338327950;
long long const MAX = 1e4 + 1;
long long mod = 1000000007;
long long n;
string s;
void solve() {
  cin >> n;
  long long a = 1, d = 0;
  for (long long i = (1), _b = (n - 1); i <= _b; i++) {
    long long x = d;
    (d = 3 * a) %= mod;
    (a = 2LL * a + x) %= mod;
  }
  cout << d << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(30);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
