#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <unordered_set>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define _rep(i, n) _rep2(i, 0, n)
#define _rep2(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(...) GET_MACRO(__VA_ARGS__, _rep2, _rep)(__VA_ARGS__)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using i64 = long long;
template<class T, class U>
bool chmin(T& a, const U& b) { return (b < a) ? (a = b, true) : false; }
template<class T, class U>
bool chmax(T& a, const U& b) { return (b > a) ? (a = b, true) : false; }
inline void YesNo(bool f) { std::cout << (f? "Yes": "No") << std::endl; }

template<typename T>istream& operator>>(istream&i,vector<T>&v){rep(j,v.size())i>>v[j];return i;}
template<typename T>string join(vector<T>&v){stringstream s;rep(i,v.size())s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,vector<T>&v){if(v.size())o<<join(v);return o;}
template<typename T>string join(vector<vector<T>>&vv){string s="\n";rep(i,vv.size())s+=join(vv[i])+"\n";return s;}
template<typename T>ostream& operator<<(ostream&o,vector<vector<T>>&vv){if(vv.size())o<<join(vv);return o;}

int dp1[10001][1001];
int dp2[10001][1001];

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    cin >> a;
    dp1[0][0] = 1;
    dp2[0][0] = 1;
    rep(i, n){
        rep(j, k + 1){
            dp1[i + 1][j] |= dp1[i][j];
            if (j + a[i] <= k) dp1[i + 1][j + a[i]] |= dp1[i][j];
        }
    }
    if (dp1[n][k] == 0){
        cout << -1 << endl;
        return 0;
    }
    reverse(all(a));
    rep(i, n){
        rep(j, k + 1){
            dp2[i + 1][j] |= dp2[i][j];
            if (j + a[i] <= k) dp2[i + 1][j + a[i]] |= dp2[i][j];
        }
    }
    int ans = 0;
    rep(i, n){
        int c = 1;
        rep(j, k + 1){
            if (dp1[i][j] & dp2[n - i - 1][k - j]) c = 0;
        }
        ans += c;
    }
    cout << ans << endl;
}