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

#pragma gcc optimize("Ofast")

#include "bits/stdc++.h"
using namespace std;
typedef int64_t lld;
typedef pair<int,int> pii;
typedef pair<lld,lld> pll;
typedef pair<int,pll> pip;
typedef pair<pll,int> ppi;
typedef pair<lld,pll> plp;
typedef pair<pll,lld> ppl;
typedef pair<pll,pll> ppp;
template<typename T>
using maxHeap = priority_queue<T,vector<T>,less<T>>;
template<typename T>
using minHeap = priority_queue<T,vector<T>,greater<T>>;
#define ff first
#define ss second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define endl '\n'
#define jizz cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
inline void input(int &_x) {
    _x = 0;
    int _tmp = 1; char _tc = getchar();
    while((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
    if(_tc == '-') _tc = getchar(), _tmp = -1;
    while(_tc >= '0' && _tc <= '9') _x = _x*10+(_tc-48), _tc = getchar();
    _x *= _tmp;
}
inline void output(int _x) {
    char _buff[20]; int _f = 0;
    if(_x == 0)putchar('0');
    while(_x > 0)
    {
        _buff[_f++] = _x%10+'0';
        _x /= 10;
    }
    for(_f-=1; _f >= 0; _f--)
        putchar(_buff[_f]);
    putchar('\n');
}
template<typename Iter>
ostream& _out(ostream &s, Iter b, Iter e) {
    s<<"[";
    for ( auto it=b; it!=e; it++ ) s<<(it==b?"":" ")<<*it;
    s<<"]";
    return s;
}
template<class T1,class T2>
ostream& operator<<(ostream& out, pair<T1,T2> p) {
    return out << '(' << p.first << ", " << p.second << ')';
}
template<typename T>
ostream& operator <<( ostream &s, const vector<T> &c ) {
     return _out(s,c.begin(),c.end());
 }
#ifdef erd1
    #define pprint(x) cerr<<__PRETTY_FUNCTION__<<":"<<__LINE__<<" - "<<(#x)<<"="<<(x)<<endl
#else
    #define pprint(x)
#endif

// code starts here
vector<int> v,nes;
vector<vector<int>> dp;
void dfs(int cur, int mn){
    for(auto i: dp[cur]){
        if(i > mn)nes[i] = true;
        dfs(cur-i,mn);
    }
    dp[cur].resize(0);
}
signed main(){
    int n, k;
    cin >> n >> k;
    dp.resize(2*k+1);
    nes.resize(k);
    dp[0].pb(0);
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        v.pb(x);
    }
    sort(all(v));
    for(auto x: v){
        if(x >= k)break;
        for(int i = k; i >= 0; i--){
            if(dp[i].size() && !(dp[i+x].size() && dp[i+x].back()==x))dp[i+x].pb(x);
        }
    }
    dp[0].resize(0);
    for(int i = 0; i <= k; i++)dfs(i+k,i);
    int ans = n;
    for(auto x: v){
        if(x >= k || nes[x])ans--;
    }
    cout << ans << endl;
}