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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int p = sc.nextInt();
        int o = 0, e = 0;
        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            if(x % 2 == 0){
                e++;
            }else{
                o++;
            }
        }
        Combination oc = new Combination(o, Integer.MAX_VALUE);
        Combination ec = new Combination(e, Integer.MAX_VALUE);
        long ans = 0;
        for (int i = p; i <= o; i+=2) {
            for (int j = 0; j <= e; j++) {
                ans += oc.comb(o, i) * ec.comb(e, j);
            }
        }
        System.out.println(ans);
        sc.close();

    }

}

class Combination {

    /** factorial:階乗 */
    private long fac[];
    /** inverse:逆元（ここでは、fac[i]をMOD-2乗したものをMODで割った余り） */
    private long inv[];
    /** 除数 */
    private int mod;

    /**
     * コンストラクタ
     * @param size 組み合わせの対象となる個数
     * @param mod 除数
     */
    public Combination(int size, int mod){
        this.fac = new long[size +1];
        this.inv = new long[size +1];
        this.mod = mod;

        this.fac[0] = 1;
        this.inv[0] = 1;

        for(int i = 1; i <= size; i++){
            fac[i] = (fac[i -1] * i) %mod;
            inv[i] = modPow(fac[i], (int)mod -2) %mod;
        }
    }

    /**
     * n個の中からr個選ぶ組み合わせの数（nCr）をmodで割った余りを算出する
     * @param n 元の集合の要素数
     * @param r 組み合わせる要素数
     * @return nCr % mod
     */
    public long comb(int n, int r){
        return fac[n] * inv[r] %mod * inv[n - r] %mod;
    }

    /**
     * base^exp をmodで割った余りを算出する
     * @param base 基数
     * @param exp 指数
     * @return base^exp % mod
     */
    private long modPow(long base, int exp) {
        long ret = 1;
        while(exp > 0){
            if((exp & 1) == 1){
                ret = ret * base %mod;
            }
            base = base * base %mod;
            exp >>= 1;
        }
        return ret;
    }

}