Problem:
Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

var N = 0
var A = 0

fun main(args: Array<String>) {
    readLine()?.split(" ")?.map {
        it.toInt()
    }?.let {
        N = it[0]
        A = it[1]
    }


    val x = Array(N + 1) { 0 }
    readLine()?.split(" ")?.map {
        it.toInt()
    }?.forEachIndexed { i, n ->
        x[i + 1] = n
    }

    val dp = Array(N + 1) {
        Array(N + 1) {
            Array(N * A + 1) { 0L }
        }
    }
    for (i in 0..N) {
        dp[i][0][0] = 1
    }

    for (i in 1..N) {
        for (j in 1..i) {
            for (k in 1..(N * A)) {
                dp[i][j][k] = if (k < x[i]) {
                    dp[i - 1][j][k]
                } else {
                    dp[i - 1][j][k] + dp[i - 1][j - 1][k - x[i]]
                }
            }
        }
    }

    var ans = 0L
    for (j in 1..N) {
        ans += dp[N][j][j * A]
    }

    println(ans)
}