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 Python (3.4.3) (judge verdict: Accepted)

from operator import mul
from functools import reduce
# from collections import Counter
# from itertools import combinations as comb
# from itertools import permutations as perm
# from copy import copy
# 配列二分法アルゴリズム
# https://docs.python.jp/3/library/bisect.html
# import bisect
# 桁数指定
# print('{:.3f}'.format(X))
# from collections import defaultdict
# dic = defaultdict(lambda: ...)
# 値で辞書をソート
# sorted(dic.items(), key=lambda x:x[1])
# ヒープキュー
# https://docs.python.org/ja/3/library/heapq.html
# import heapq
# 正規表現(regular expression)のためのモジュール
# import re

import sys

sys.setrecursionlimit(2000)


def inpl():
    return list(map(int, input().split()))


# 重複組み合わせは nHr = (n + r - 1)Cn
def cmb(n, r):
    # combination
    if n < r:
        return 0
    r = min(n - r, r)
    if r == 0:
        return 1
    over = reduce(mul, range(n, n - r, -1))
    under = reduce(mul, range(1, r + 1))
    return over // under


def gcd(a, b):
    # greatest common divisor
    while b > 0:
        a, b = b, a % b

    return a


def lcm(a, b):
    # least common multiple
    return a * b // gcd(a, b)


N, P = inpl()
A = inpl()

odd = sum(a % 2 == 1 for a in A)
even = sum(a % 2 == 0 for a in A)

if P == 1 and odd == 0:
    print(0)
    exit()

print(
    sum(cmb(odd, i) for i in range(P, odd + 1, 2)) * sum(
        cmb(even, j) for j in range(even + 1)))