Problem:
Score : 600 points

Problem Statement

AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i.
Because he loves big numbers, he calls a subset of the cards good when the sum of the numbers written on the cards in the subset, is K or greater.

Then, for each card i, he judges whether it is unnecessary or not, as follows:

If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary.

Otherwise, card i is NOT unnecessary.

Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.

Constraints

All input values are integers.

1≤N≤5000

1≤K≤5000

1≤a_i≤10^9 (1≤i≤N)

Partial Score

300 points will be awarded for passing the test set satisfying N,K≤400.

Input

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

N K
a_1 a_2 ... a_N

Output

Print the number of the unnecessary cards.

Sample Input 1

3 6
1 4 3

Sample Output 1

1

There are two good sets: {2,3} and {1,2,3}.

Card 1 is only contained in {1,2,3}, and this set without card 1, {2,3}, is also good. Thus, card 1 is unnecessary.

For card 2, a good set {2,3} without card 2, {3}, is not good. Thus, card 2 is NOT unnecessary.

Neither is card 3 for a similar reason, hence the answer is 1.

Sample Input 2

5 400
3 1 4 1 5

Sample Output 2

5

In this case, there is no good set. Therefore, all the cards are unnecessary.

Sample Input 3

6 20
10 4 3 10 25 2

Sample Output 3

3

Solution in PyPy2 (5.6.0) (judge verdict: Memory Limit Exceeded)

def main():
    def accumulate(X):
        ret = [X[0]]
        for x in X[1:]:
            ret.append(ret[-1] + x)
        return ret
    
    N, K = map(int, raw_input().split())
    A = list(map(int, raw_input().split()))

    # K以上の数はすべて必要なのでA未満の数だけを考える
    A = sorted([a for a in A if a < K], reverse=True)
    N = len(A)

    # dp1[i][k] := 1 ~ iまでのカードを使って数kが作れるか
    dp1 = [[0] * (K + 1) for i in range(N + 1)]
    dp1[0][0] = 1
    # dp2[i][k] := N ~ iまでのカードを使って数kが作れるか
    dp2 = [[0] * (K + 1) for i in range(N + 1)]
    dp2[0][0] = 1

    # dpテーブルを計算
    for i in range(1, N + 1):
        for k in range(K + 1):
            dp1[i][k] = max(dp1[i][k], dp1[i - 1][k])
            if A[i - 1] <= k:
                dp1[i][k] = max(dp1[i][k], dp1[i - 1][k - A[i - 1]])

            dp2[i][k] = max(dp2[i][k], dp2[i - 1][k])
            if A[N - i] <= k:
                dp2[i][k] = max(dp2[i][k], dp2[i - 1][k - A[N - i]])

    # 累積和を取っておく
    for i in range(N + 1):
        dp2[i] = list(accumulate(dp2[i]))

    # ansを計算
    need = 0
    for i in range(1, N + 1):
        for k in range(K + 1):
            # 1 ~ i番目まででkが作れない
            if not dp1[i - 1][k]:
                continue

            # 1 ~ i番目までだけで必要な条件が作れる
            if K - A[i - 1] <= k < K:
                need += 1
                break

            # 範囲外参照
            if K - A[i - 1] - k - 1 < 0:
                continue

            if dp2[N - i][K - k] - dp2[N - i][K - A[i - 1] - k - 1] > 0:
                need += 1
                break

    print(N - need)

main()