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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int,int>;
using pl4 = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define REP(i,a) for(int i=0;i<(a);i++)
#define REP1(i,a) for(int i=1;i<=(a);i++)
#define rREP(i,a) for(int i=(a-1);i>=0;i--)
#define rREP1(i,a) for(int i=(a);i>0;i--)
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))

#define int ll

const ll MOD = 1e9+7;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll sgn(ll n){
    if(n==0) return 0;
    else return n/abs(n);
}

ll salv(vll v){
  say("{");
  FOR(i,0,v.sz-1){
    say(v[i]);
    if(i!=v.sz-1) say(",");
  }
  sal("}")
}

ll DigS10(ll n){
  ll m=0;
  FOR(i,0,DigN10(n)-1){
    m+=(ll)((llabs(n)%(ll)(pow(10.0,(double)(i+1))))/(ll)(pow(10.0,(double)i)));
  }
  return m;
}

ll isP(ll n){
  if(n<=1) return 0;
  FOR(i,2,(ll)sqrt(n)){
    if(n%i==0) return 0;
  }
  return 1;
}

vll FactM(1,1);
vll FactMI(1,1);

ll PowM(ll a,ll b){
  ll ans=1,x=(a%MOD);
  FOR(i,0,DigN2(b)-1){
    if(Dig2(b,i)==1) ans=(ans*x)%MOD;
    if(i!=(DigN2(b)-1)) x=(x*x)%MOD;
  }
  return ans;
}

void CFactM(ll n){
  FOR(i,FactM.sz,n){
    FactM.pb((FactM[i-1]*(i%MOD))%MOD);
  }
  return;
}

void CFactMI(ll n){
  CFactM(n);
  if(FactMI.sz<(n+1)) FactMI.res(n+1,-1);
  if(FactMI[n]==-1) FactMI[n]=PowM(FactM[n],MOD-2);
  rFOR(i,1,n-1){
    if(FactMI[i]!=-1) break;
    FactMI[i]=((FactMI[i+1]*((i+1)%MOD))%MOD);
  }
  return;
}

ll CombM(ll n,ll k){
    if((n<0)||(k<0)) return 0;
    if(n<k) return 0;
    if(n+1>FactMI.sz) CFactMI(n);
    return ((((FactMI[k]*FactMI[n-k])%MOD)*FactM[n])%MOD);
}

signed main() {
  ll n,p,a,r=0,m;
  cin >> n >> p;
  REP(i,n){
    cin >> a;
    if(a%2)r++;
  }
  if(!r){
    if(p){
      cout << 0;
    } else {
      cout << (ll)pow(2,n);
    }
    return 0;
  }
  cout << (ll)pow(2,n-1) << endl;
  return 0;
}