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 PHP (7.4.4) (judge verdict: Accepted)

<?php
define("MOD", 10 ** 9 + 7);
list($n, $p) = ints();
$a = ints();
$cntNoun = $cntOdd = 0;
for ($i = 0; $i < $n; ++$i) {
    if ($a[$i] % 2) ++$cntOdd;
    else ++$cntNoun;
}
// 偶数個の袋は選び方は何でも良い
$k = 2 ** $cntNoun;
init($n);
if ($p) {
    // 奇数個の袋は奇数個選ぶ
    $t = 0;
    for ($i = 1; $i <= $cntOdd; $i += 2) $t += nCr2($cntOdd, $i);
} else {
    // 奇数個の袋は偶数個選ぶ
    $t = 0;
    for ($i = 0; $i <= $cntOdd; $i += 2) $t += nCr2($cntOdd, $i);
}
echo $k * $t;
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}
// 足し算
function modAdd($x, $y)
{
    return ($x + $y) % MOD;
}

// 引き算
function modSub($x, $y)
{
    return ($x + MOD - $y) % MOD;
}

// 掛け算
function modMul($x, $y)
{
    return ($x * $y) % MOD;
}

// 割り算
function modDiv($x, $y)
{
    return modMul($x, modPow($y, MOD - 2));
}

// 累乗（繰り返し二乗法）
function modPow($n, $x)
{
    if ($x === 0) return 1;
    $res = (modPow($n, $x >> 1) ** 2) % MOD;
    if ($x % 2 === 1) $res = modMul($res, $n);
    return $res;
}

// 階乗
function modFac($n)
{
    if ($n === 0) return 1;
    return modMul($n, modFac($n - 1));
}

// 順列
function nPr($n, $r)
{
    if ($r === 0) return 1;
    return modMul(nPr($n, $r - 1), $n - $r + 1);
}

// 組み合わせ
function nCr($n, $r)
{
    $r = min($r, $n - $r);
    if ($r === 0) return 1;
    return modDiv(nPr($n, $r), modFac($r));
}

// 階乗、逆元を前処理
function init($n)
{
    global $fact, $ifact;
    $fact[0] = 1;
    for ($i = 1; $i <= $n; ++$i) $fact[$i] = modMul($fact[$i - 1], $i);
    $ifact[$n] = modDiv(1, $fact[$n]);
    for ($i = $n; $i >= 1; --$i)
        $ifact[$i - 1] = modMul($ifact[$i], $i);
}

// 順列（前処理）
function nPr2($n, $r)
{
    global $fact, $ifact;
    return modMul($fact[$n], $ifact[$n - $r]);
}

// 組み合わせ（前処理）
function nCr2($n, $r)
{
    global $fact, $ifact;
    return modMul(modMul($fact[$n], $ifact[$r]), $ifact[$n - $r]);
}