You have a given integer n. Find the number of ways to fill all 3 × n tiles with the shape described in the picture below. Upon filling, no empty spaces are allowed. Shapes cannot overlap.

<image> This picture describes when n = 4. The left one is the shape and the right one is 3 × n tiles. 

Input

The only line contains one integer n (1 ≤ n ≤ 60) — the length.

Output

Print the number of ways to fill.

Examples

Input


4


Output


4

Input


1


Output


0

Note

In the first example, there are 4 possible cases of filling.

In the second example, you cannot fill the shapes in 3 × 1 tiles.
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
const long double PI = 2 * acos(0.0);
const vector<long long int> dx = {1, -1, 0, 0};
const vector<long long int> dy = {0, 0, 1, -1};
vector<long long int> ga(long long int n, bool oneIndexed = false) {
  vector<long long int> a;
  if (oneIndexed) a.push_back(0ll);
  for (long long int i = 0; i < n; i++) {
    long long int p;
    cin >> p;
    a.push_back(p);
  }
  return move(a);
}
vector<unsigned long long int> gau(unsigned long long int n,
                                   bool oneIndexed = false) {
  vector<unsigned long long int> a;
  if (oneIndexed) a.push_back(0llu);
  for (unsigned long long int i = 0; i < n; i++) {
    unsigned long long int p;
    cin >> p;
    a.push_back(p);
  }
  return move(a);
}
vector<string> gas(unsigned long long int n, bool oneIndexed = false) {
  vector<string> a;
  if (oneIndexed) a.push_back("");
  for (unsigned long long int i = 0; i < n; i++) {
    string p;
    cin >> p;
    a.push_back(p);
  }
  return move(a);
}
template <typename T, typename A>
void pa(vector<T, A> const &a, long long int begin = 0,
        long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
template <typename T, typename A>
void papair(vector<T, A> const &a, long long int begin = 0,
            long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
  cout << "\n";
}
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
void yesno(int f) {
  if (f)
    yes();
  else
    no();
}
void solve() {
  long long int n;
  cin >> n;
  long long int blocks = n / 2;
  if (n & 1)
    cout << 0;
  else
    cout << (long long int)pow(2, blocks);
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
