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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
inline long long readl() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int power(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    b >>= 1;
    a = a * a;
  }
  return ans;
}
int power_mod(int a, int b, int mod) {
  a %= mod;
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b >>= 1, a = (a * a) % mod;
  }
  return ans;
}
long long powerl(long long a, long long b) {
  long long ans = 1ll;
  while (b) {
    if (b & 1ll) ans = ans * a;
    b >>= 1ll;
    a = a * a;
  }
  return ans;
}
long long power_modl(long long a, long long b, long long mod) {
  a %= mod;
  long long ans = 1ll;
  while (b) {
    if (b & 1ll) ans = (ans * a) % mod;
    b >>= 1ll, a = (a * a) % mod;
  }
  return ans;
}
long long gcdl(long long a, long long b) { return b == 0 ? a : gcdl(b, a % b); }
long long abssl(long long a) { return a > 0 ? a : -a; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int abss(int a) { return a > 0 ? a : -a; }
long long n, dp[65];
int main() {
  cin >> n;
  dp[0] = 1;
  for (long long i = 2; i <= n; ++i) {
    dp[i] += dp[i - 2] * 2;
  }
  printf("%lld\n", dp[n]);
  return 0;
}
