#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define endl "\n"
#define rep(i, a, b) for (auto (i) = (a); (i) < (b); (i)++)
#define rrep(i, a, b) for (auto (i) = (a); (i) > (b); (i)--)

#ifdef LOCAL_DEFINE
struct InitInput{
  InitInput() {
    FILE *stream1;
    stream1=freopen("/Users/aim_cpo/CLionProjects/competitive/in.txt","r",stdin);
    assert(stream1 != nullptr);
    cerr << "This problem is not interactive" << endl;
  }
}LOCAL_INPUT;
struct LOCAL_OUTPUT{
  LOCAL_OUTPUT() {
    FILE *stream2;
    const char *outputfile = "out.txt";
    stream2 = freopen(outputfile, "w", stdout);
    assert(stream2 != nullptr);
    cerr << "output [ " << outputfile << " ]" << endl;
  }
}/*LOCAL_OUTPUT*/;
#define show(x) cerr << #x << " = " << (x) << " (line " << __LINE__ << ")" << endl
#define showV(v, n) do{\
for(int _i_t_e_=0; _i_t_e_<(n);_i_t_e_++)\
cerr<<"("<<_i_t_e_<<" = "<<(v)[_i_t_e_]<<") ";\
cerr<<endl;}while(0)
#define showA(a, n) do{\
for(int _i_t_e_=0;_i_t_e_<(n);_i_t_e_++)\
cerr<<"("<<_i_t_e_<<" = "<<(a)[_i_t_e_]<<") ";\
cerr<<endl;}while(0)
#define showA2(a, n, m) do{\
for(int _i_t_e_=0;_i_t_e_<(n);_i_t_e_++){\
for(int _i_t_e_2=0;_i_t_e_2<(m);_i_t_e_2++){\
cerr<<"("<<_i_t_e_<<", "<<_i_t_e_2<<" = "<<(a)[_i_t_e_][_i_t_e_2]<<") ";\
}cerr<<endl;}cerr<<endl;}while(0)
#else
#define show(x)
#define showV(v, n)
#define showA(a, n)
#define showA2(a, n, m)
#endif

struct InitAim{
  InitAim() {
    ios::sync_with_stdio(false);cin.tie(nullptr);
    cout.precision(12);cout<<fixed;
#ifdef LOCAL_DEFINE
    cerr << "This problem is not interactive" << endl;
#endif
  }
}aim_cpo;

typedef long long ll;
typedef unsigned long long ull;

constexpr const int INT_INF=0x3f3f3f3f; //1061109567
constexpr const ll LL_INF=0x3f3f3f3f3f3f3f3f; //4557430888798830399

template <typename T> bool chmin(T &a, T b){return a>b?(a=b,true):false;}
template <typename T> bool chmax(T &a, T b){return a<b?(a=b,true):false;}
template <typename T> void ln(T i, T n){cout<<(i==n-1?"\n":" ");}
template<typename T, typename S>
ostream &operator<<(ostream &out, const pair<T, S> &pair1) {
  out << '(' << pair1.fi << ", " << pair1.se << ')';
  return out;
}
template<typename T>
ostream &operator<<(ostream &out, const vector<T> &vector1) {
  out << '[';
  for (unsigned int i = 0; i < vector1.size(); i++) {
    out << vector1[i];
    if (i == vector1.size() - 1)out << "]"; else out << ", ";
  }
  return out;
}

// INT
#define GCD(a, b) __gcd(a, b)
template <typename T> T LCM(T a, T b) {return a / GCD(a, b) * b;}
template <typename T> T EXTGCD(T a, T b, T& x, T& y) {
  T d = a;
  if (b != 0) {d=EXTGCD(b,a%b,y,x);y-=(a/b)*x;}
  else x=1,y=0;
  return d;
}
template <typename T> bool is_prime(T a) {
  for(int i=2;i*i<=a;i++)if(a%i==0)return 1;
  return 0;
}
template <typename T, typename S> T Pow(T a, S b) {
  T res = 1, now = a;
  while(b){if(b&1)res*=now;b>>=1;now*=now;}return res;
}

// MOD
ll MOD = 1000000000L + 7L;
#define add(a, b) ((a % MOD) + (b % MOD)) % MOD
#define mul(a, b) ((a % MOD) * (b % MOD)) % MOD
#define sub(a, b) ((a % MOD) + MOD - (b % MOD)) % MOD
template <typename T, typename S> T ModPow(T a, S b) {
  T res = 1, now = a;
  while(b){if(b&1)res=mul(res,now);b>>=1;now=mul(now,now);}return res;
}
template <typename T> T mod_inverse(T a, T mod, bool prime){ // if mod is prime, "prime" is true.
  if(prime)return ModPow(a, mod - 2);
  else{T x,y;EXTGCD(a,mod,x,y);return (mod+x%mod)%mod;}
}
#define divide(a, b) ((a % MOD) * (mod_inverse(b, MOD, true))) % MOD

//LLの数値をつかう時は最後にLをつける癖をつけよう
//
//  ┏┓┏┓　┓　┏┓
//  ┏┛┃┃　┃　┗┫
//  ┗┛┗┛　┻　┗┛
//
//   謹┃賀┃新┃年┃
//   ━┛━┛━┛━┛
//WWWWWWWWWWWWWWWWWWWWWW

int n, k;
ll a[5001];
bool dp1[5001][5001];
bool dp2[5001][5001];
int sum[5001];

inline int GetSum(int left, int right) {
  if (right < 0) return 0;
  if (left < 0) left = 0;
  if (left == 0) return sum[right];
  return sum[right] - sum[left - 1];
}

int main() {
#ifdef LOCAL_DEFINE
  show(MOD);
#endif
  // テキトー考察
  //
  // a[i]以外の数で実現する事ができる和が高速に求まれば、良いのだけど...
  // ということで、解説pdfのdpを前からと後ろからやるといい事を知る(頭いいなあ)
  // そうすると、事前に2回dpしておけば、後々dpする必要なくて計算量が落ちる.
  // あとは、a[i]が必要であるならば、k - a[i]からk - 1までの数のうち最低一つが
  // a[i]以外の数で実現されていれば必要、そうじゃなきゃ必要ないので
  // 累積和使って実現できるか否か調べれば良い.
  //
  // 前からと後ろからのdpって知ったとき、感動したわ.
  // たしかに、これならある数を使わずに実現できる数を速く知ることができるよね.
  // いやー、この問題最高.好き.
  cin >> n >> k;
  rep(i, 0, n) cin >> a[i];
  dp1[0][0] = true;
  rep(i, 0, n) {
    rep(j, 0, k + 1) {
      if (!dp1[i][j]) continue;
      if (j + a[i] <= k) dp1[i + 1][j + a[i]] = true;
      dp1[i + 1][j] = true;
    }
  }
  dp2[n][0] = true;
  rrep(i, n, 0) {
    rep(j, 0, k + 1) {
      if (!dp2[i][j]) continue;
      if (j + a[i - 1] <= k) dp2[i - 1][j + a[i - 1]] = true;
      dp2[i - 1][j] = true;
    }
  }
  int ans = 0;
  rep(i, 0, n) {
    fill(sum, sum + 5001, 0);
    rep(j, 0, k + 1) if (dp2[i + 1][j]) sum[j]++;
    rep(j, 1, k + 1) sum[j] = sum[j] + sum[j - 1];
    int left = k - a[i];
    int right = k - 1;
    bool need = false;
    rep(j, 0, k + 1) {
      if (dp1[i][j]) {
        if (GetSum(left - j, right - j) > 0) {
          need = true;
          break;
        }
      }
    }
    if (!need) ans++;
  }
  cout << ans << endl;
#ifdef  LOCAL_DEFINE
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;
#endif
  return 0;
}

Measured memory_bytes: 49152