output_description: Print the number of ways to select bags so that the total number of biscuits
  inside is congruent to P modulo 2.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N P
      A_1 A_2 ... A_N
attempt: import sys
  import math
  import copy
  import heapq
  from functools import cmp_to_key
  from bisect import bisect_left, bisect_right
  from collections import defaultdict, deque, Counter
  
  # sys.setrecursionlimit(1000000)
  
  # input aliases
  input = sys.stdin.readline
  getS = lambda: input().strip()
  getN = lambda: int(input())
  getList = lambda: list(map(int, input().split()))
  getZList = lambda: [int(x) - 1 for x in input().split()]
  
  INF = float("inf")
  MOD = 10**9 + 7
  divide = lambda x: pow(x, MOD - 2, MOD)
  
  
  def nck(n, k, kaijyo):
      return (npk(n, k, kaijyo) * divide(kaijyo[k])) % MOD
  
  
  def npk(n, k, kaijyo):
      if k == 0 or k == n:
          return n % MOD
      return (kaijyo[n] * divide(kaijyo[n - k])) % MOD
  
  
  def fact_and_inv(SIZE):
      inv = [0] * SIZE  # inv[j] = j^{-1} mod MOD
      fac = [0] * SIZE  # fac[j] = j! mod MOD
      finv = [0] * SIZE  # finv[j] = (j!)^{-1} mod MOD
      inv[1] = 1
      fac[0] = fac[1] = 1
      finv[0] = finv[1] = 1
      for i in range(2, SIZE):
          inv[i] = MOD - (MOD // i) * inv[MOD % i] % MOD
          fac[i] = fac[i - 1] * i % MOD
          finv[i] = finv[i - 1] * inv[i] % MOD
  
      return fac, finv
  
  
  def renritsu(A, Y):
      # example 2x + y = 3, x + 3y = 4
      # A = [[2,1], [1,3]])
      # Y = [[3],[4]] または [3,4]
      A = np.matrix(A)
      Y = np.matrix(Y)
      Y = np.reshape(Y, (-1, 1))
      X = np.linalg.solve(A, Y)
  
      # [1.0, 1.0]
      return X.flatten().tolist()[0]
  
  
  class TwoDimGrid:
      # 2次元座標 -> 1次元
      def __init__(self, h, w, wall="#"):
          self.h = h
          self.w = w
          self.size = (h + 2) * (w + 2)
          self.wall = wall
          self.get_grid()
          # self.init_cost()
  
      def get_grid(self):
          grid = [self.wall * (self.w + 2)]
          for i in range(self.h):
              grid.append(self.wall + getS() + self.wall)
  
          grid.append(self.wall * (self.w + 2))
          self.grid = grid
  
      def init_cost(self):
          self.cost = [INF] * self.size
  
      def pos(self, x, y):
          # 壁も含めて0-indexed 元々の座標だけ考えると1-indexed
          return y * (self.w + 2) + x
  
      def getgrid(self, x, y):
          return self.grid[y][x]
  
      def get(self, x, y):
          return self.cost[self.pos(x, y)]
  
      def set(self, x, y, v):
          self.cost[self.pos(x, y)] = v
          return
  
      def show(self):
          for i in range(self.h + 2):
              print(self.cost[(self.w + 2) * i : (self.w + 2) * (i + 1)])
  
      def showsome(self, tgt):
          for t in tgt:
              print(t)
          return
  
      def showsomejoin(self, tgt):
          for t in tgt:
              print("".join(t))
          return
  
      def search(self):
          grid = self.grid
          move = [(0, 1), (0, -1), (1, 0), (-1, 0)]
          move_eight = [
              (0, 1),
              (0, -1),
              (1, 0),
              (-1, 0),
              (1, 1),
              (1, -1),
              (-1, 1),
              (-1, -1),
          ]
          # for i in range(1, self.h+1):
          #     for j in range(1, self.w+1):
          #         cx, cy = j, i
          #         for dx, dy in move_eight:
          #             nx, ny = dx + cx, dy + cy
  
  
  def judge(mid, cnt, n):
      xx = cnt[0]
      if ((mid + 1) * (xx - 1)) >= n:
          return False
      elif (mid + 1) * xx <= n:
          return True
      yoyu = n % (mid + 1)
      if Counter(cnt)[xx] <= yoyu:
          return True
      else:
          return False
  
  
  def solve():
      n, k = getList()
      nums = getList()
      for num in nums:
          if num % 2 == 1:
              print(pow(2, n - 1))
              return
      if k == 0:
          print(pow(2, n))
      else:
          print(0)
  
  
  def main():
      n = getN()
      for _ in range(n):
          solve()
  
      return
  
  
  if __name__ == "__main__":
      # main()
      solve()
  
problem_description: 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?
samples: [{"input": "2 0\n    1 3", "output": "2\n    \n\nThere are two ways to select bags so that the total number of biscuits inside\nis congruent to 0 modulo 2:\n\n  * Select neither bag. The total number of biscuits is 0.\n  * Select both bags. The total number of biscuits is 4.\n\n* * *"}, {"input": "1 1\n    50", "output": "0\n    \n\n* * *"}, {"input": "3 0\n    1 1 1", "output": "4\n    \n\nTwo bags are distinguished even if they contain the same number of biscuits.\n\n* * *"}, {"input": "45 1\n    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", "output": "17592186044416"}]