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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;

namespace AtCoder.Grand017
{
    public class AGC017A_Biscuits
    {
        public static void Main(string[] args)
        {
            
            // 総数と余りの数の入力
            int[] nums = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            int n = nums[0];
            int p = nums[1];

            // 袋毎のビスケット数の入力
            int[] biscuitNum = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();

            int evenNum = biscuitNum.Count(num => num % 2 == 0);
            int oddNum = biscuitNum.Count(num => num % 2 == 1);

            if(p == 0)
            {
                // 偶数枚食べる
                // 偶数袋から任意の数と、奇数袋から偶数
                
                long evenCombi = 0;
                for(int i=0; i <= evenNum; i++)
                {
                    evenCombi += MathMod.Combination(evenNum, i);
                }

                long oddCombi = 0;
                for(int i = 0; i <= oddNum; i+= 2)
                {
                    oddCombi += MathMod.Combination(oddNum, i);
                }

                // 結果出力
                Console.WriteLine(evenCombi * oddCombi);

            }
            else
            {
                // 奇数枚食べる
                // 偶数袋から任意の数と、奇数袋から奇数

                long evenCombi = 0;
                for(int i = 0; i <= evenNum; i++)
                {
                    evenCombi += MathMod.Combination(evenNum, i);
                }

                long oddCombi = 0;
                for(int i = 1; i <= oddNum; i += 2)
                {
                    oddCombi += MathMod.Combination(oddNum, i);
                }

                // 結果出力
                Console.WriteLine(evenCombi * oddCombi);

            }
            

        }


        public class MathMod
        {

            // 階乗計算に使う素数(階乗計算の上限数) 10^9+7
            static int modVal = 1000000007;

            // 階乗計算を行う数
            private static int FactorialMax = 60;


            // 階乗の剰余
            private static int[] factorialMod = new int[FactorialMax + 1];

            // 階乗の逆元
            private static int[] factorialInverse = new int[FactorialMax + 1];


            static MathMod()
            {

                // 階乗の剰余・逆元を計算
                factorialMod[0] = 1;
                factorialInverse[0] = 1;
                for(long i = 1; i <= FactorialMax; i++)
                {
                    factorialMod[i] = (int)((factorialMod[i - 1] * i) % modVal);
                    factorialInverse[i] = Pow(factorialMod[i], modVal - 2);
                }

            }

            /// <summary>累乗を求める</summary>
            public static int Pow(int x, int y)
            {
                long remaining = y;
                long p = x;

                long ret = 1;

                while(remaining > 0)
                {
                    if(remaining % 2 == 0)
                    {
                        p = (p * p) % modVal;
                        remaining /= 2;
                    }
                    else
                    {
                        ret = (ret * p) % modVal;
                        remaining -= 1;
                    }
                }

                return (int)ret;
            }


            /// <summary>階乗を求める</summary>
            public static double Factorial(int n)
            {
                if(n < 0 || n > FactorialMax)
                {
                    throw new ArgumentOutOfRangeException();
                }

                return factorialMod[n];
            }


            /// <summary>組み合わせ数を求める</summary>
            /// <param name="n">全体集合の要素数</param>
            /// <param name="r">取り出す要素数</param>
            /// <returns>組み合わせ数</returns>
            public static int Combination(int n, int r)
            {
                if(n > FactorialMax || r > FactorialMax)
                {
                    throw new ArgumentOutOfRangeException();
                }

                if(n < r || n < 0 || r < 0)
                {
                    return 0;
                }

                return (int)((long)factorialMod[n] * factorialInverse[r] % modVal * factorialInverse[n - r] % modVal);
            }

        }

    }
}