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>
const long long INF = (1LL << 32) - 1;
const long long INF_LL = 9223372036854775807LL;
const double E = exp(1.0);
const double PI = acos(-1.0);
long long gcd(long long a, long long b) {
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
long long lcd(long long a, long long b) { return a * b / gcd(a, b); }
long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void init_cin() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
}
using namespace std;
int main() {
  int n = read();
  if (n % 2 == 0) {
    printf("%I64d\n", (long long)pow(2, n / 2));
  } else {
    puts("0");
  }
}
