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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define rep(i,N) for(int i = 0; i < (N); i++)
#define erep(i,N) for(int i = N; i >= 0; i--)
const ll INF = 1000000007;
#define PI 3.1415926535
using Graph = vector<vector<int>>;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true;} return false;}
bool seen[100000];
void dfs(const Graph &G, int v) {
    seen[v] = true; // v を訪問済にする
 
    // v から行ける各頂点 next_v について
    for (auto next_v : G[v]) { 
        if (seen[next_v]) continue; // next_v が探索済だったらスルー
        dfs(G, next_v); // 再帰的に探索
    }
}
 
//dpTable
ll dp[55][2];
int main(){
    int N,P;
    cin >> N >> P;
    vector<int> A(N);
    rep(i,N) cin >> A.at(i);
    dp[0][0] = 1;
    for (int i = 1; i <= N; i++) {
        if (A.at(i - 1) % 2 == 0) {
            dp[i][0] = dp[i - 1][0] * 2;
            dp[i][1] = dp[i - 1][1] * 2;
        }
        else {
            dp[i][0] = dp[i - 1][0] + dp[i - 1][1];
            dp[i][1] = dp[i - 1][0] + dp[i - 1][1];
        }
    }
    cout << dp[N][P] << endl;
    return 0;
}