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;
bool isPrime[1000005] = {false};
vector<long long> prm;
vector<long long> SPF(1000005);
void sieve() {
  isPrime[1] = isPrime[0] = 0;
  long long i;
  SPF[1] = SPF[0] = 1;
  for (i = 2; i < 1000005; ++i) isPrime[i] = 1, SPF[i] = i;
  for (i = 2; i * i < 1000005; ++i) {
    if (isPrime[i]) {
      for (long long j = i * i; j < 1000005; j += i) {
        isPrime[j] = 0;
        if (SPF[j] == j) SPF[j] = i;
      }
    }
  }
  if (!prm.empty()) prm.clear();
  for (i = 2; i < 1000005; ++i) {
    if (isPrime[i]) prm.push_back(i);
  }
}
void printa(bool f) {
  if (f)
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
}
bool flip(bool f) {
  if (f) return 0;
  return 1;
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long prod = 1;
  while (b > 0) {
    if (b & 1) prod = prod * a;
    b = b >> 1;
    a = a * a;
  }
  return prod;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res *= i;
  }
  return res;
}
long long nCr(long long n, long long r) {
  return (long double)fact(n) / (long double)(fact(r) * fact(n - r));
}
void solve() {
  long long n;
  cin >> n;
  if (n < 2 || (n & 1))
    cout << 0 << "\n";
  else if (!(n & 1))
    cout << fastPower(2, n / 2) << "\n";
}
int main() {
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
