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;
long long power(long long x, long long y);
bool isPrime(long long n);
vector<long long> X[200001], Y[200001];
long long A[200001], B[200001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b, c, d, n, m, i, j, k, l, p, q, x, y, z, cnt = 0, flag = 0;
  string s1, s2, s3;
  vector<long long> V, W;
  vector<long long>::iterator itrv;
  map<long long, long long> M;
  set<long long> S;
  set<long long>::iterator itrs;
  cin >> n;
  if (n % 2 != 0) {
    cout << "0";
    return 0;
  }
  p = pow(2, n / 2);
  cout << p;
  return 0;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long p = sqrt(n);
  for (int i = 5; i <= p; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
