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 Python (3.4.3) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-

import sys
import numpy as np

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

def bisearch_min(mn, mx, func):
    """ 条件を満たす最小値を見つける二分探索 """

    ok = mx
    ng = mn
    while ng+1 < ok:
        mid = (ok+ng) // 2
        if func(mid):
            # 下を探しに行く
            ok = mid
        else:
            # 上を探しに行く
            ng = mid
    return ok

N, K = MAP()
A = LIST()

def check(omit):
    # 部分和DP
    dp = np.zeros(K, dtype=np.bool)
    dp[0] = 1
    for i in range(N):
        if i != omit and A[i] < K:
            # まとめて遷移
            dp[A[i]:] |= dp[:K-A[i]]
    # 今回の値A[omit]を足してKに到達できるような部分和があれば、これは必要
    for j in range(max(0, K-A[omit]), K):
        if dp[j]:
            return True
    # なければ不要
    return False

A.sort()
print(bisearch_min(-1, N, check))