Problem:
Score : 200 points

Problem Statement

There are N bags of biscuits. The i-th bag contains A_i biscuits.

Takaki will select some of these bags and eat all of the biscuits inside.
Here, it is also possible to select all or none of the bags.

He would like to select bags so that the total number of biscuits inside is congruent to P modulo 2.
How many such ways to select bags there are?

Constraints

1 \leq N \leq 50

P = 0 or 1

1 \leq A_i \leq 100

Input

Input is given from Standard Input in the following format:

N P
A_1 A_2 ... A_N

Output

Print the number of ways to select bags so that the total number of biscuits inside is congruent to P modulo 2.

Sample Input 1

2 0
1 3

Sample Output 1

2

There are two ways to select bags so that the total number of biscuits inside is congruent to 0 modulo 2:

Select neither bag. The total number of biscuits is 0.

Select both bags. The total number of biscuits is 4.

Sample Input 2

1 1
50

Sample Output 2

0

Sample Input 3

3 0
1 1 1

Sample Output 3

4

Two bags are distinguished even if they contain the same number of biscuits.

Sample Input 4

45 1
17 55 85 55 74 20 90 67 40 70 39 89 91 50 16 24 14 43 24 66 25 9 89 71 41 16 53 13 61 15 85 72 62 67 42 26 36 66 4 87 59 91 4 25 26

Sample Output 4

17592186044416

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define BIT(n) (1LL << (n))
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPB(i, n) for (int i = 0; i < BIT(n); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORA(i, n) for (auto &&i : n)
#define FORS(i, m, n) for (int i = m; i <= n; i++)
#define DEBUG(x) cerr << #x << ": " << x << "\n"
#define DEBUG2(x, y) cerr << #x << ": " << x << " " << #y << ": " << y << "\n"
#define DEBUG3(x, y, z) cerr << #x << ": " << x << " " << #y << ": " << y << " " << #z << ": " << z << "\n"
#define PRINTALL(V)    \
  for (auto v : (V)) { \
    cerr << v << " ";  \
  }                    \
  cerr << "\n";
#define ALL(v) v.begin(), v.end()
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define epb emplace_back
#define int long long
#define pint pair<int, int>
#define ld long double
using namespace std;
template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> using vec = std::vector<T>;
template <class T> void print(const T &x) { cout << x << "\n"; }
const int MOD = 1000000007, INF = 1061109567, INF2 = INF * INF;
const double EPS = 1e-10, PI = acos(-1.0);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const int CALCSIZE = 5010;
int C[CALCSIZE][CALCSIZE];

void init() {
  for (int i = 0; i < CALCSIZE; ++i) {
    C[i][i] = C[i][0] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
  }
}
signed main() {
  cin.tie(0), ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  init();
  int N, P;
  cin >> N >> P;
  int cnt = 0;
  REP(i, N) {
    int q;
    cin >> q;
    cnt += q % 2 == 1;
  }
  int val = 0;
  FORS(i, 0, cnt) {
    if (i % 2 == P) val += C[cnt][i];
  }

  print(BIT(N - cnt) * val);
}