#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;

#define rep(i, n) for(ll i = 0; i < n; i++)
#define rep2(i, m, n) for(ll i = m; i <= n; i++)
#define rrep(i, m, n) for(ll i = m; i >= n; i--)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define SZ(a) ((ll)(a).size())
#define bit(n, k) ((n >> k) & 1)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const int dx[4] = { -1, 1, 0, 0 };
const int dy[4] = { 0, 0, 1, -1 };
const int inf = 1e9 + 1;
const ll INF = 1e18;
const double pi = acos(-1);

using Graph = vector<vector<ll>>;
using REV_PQ = priority_queue<ll, vector<ll>, greater<ll>>;
using PQ = priority_queue<ll>;

//typedef atcoder::modint998244353 mint;
typedef atcoder::modint1000000007 mint;

const int SIZE = 200010;

ll powll(ll n, ll x) {
    // return n ^ x;

    ll ret = 1;
    rep(i, x) ret *= n;

    return ret;
}

/*struct Edge { ll u, v, cost; };

bool comp(const Edge& e1, const Edge& e2) {
    return e1.cost < e2.cost;
}*/

void OutputYesNo(bool val) {

    if (val) cout << "Yes" << endl;
    else cout << "No" << endl;
}
void OutputTakahashiAoki(bool val) {

    if (val) cout << "Takahashi" << endl;
    else cout << "Aoki" << endl;
}

void OutPutInteger(ll x) { cout << x << endl; }
void OutPutString(string x) { cout << x << endl; }


bool dp1[10010][1010];
bool dp2[10010][1010];
bool seen[10010];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, k; cin >> n >> k;
    vector<int> a(n);
    rep(i, n) cin >> a[i];

    dp1[0][0] = true;
    dp2[0][0] = true;

    rep(i, n)rep2(j, 0, k) {
        if (j - a[i] >= 0) {
            dp1[i + 1][j] |= dp1[i][j - a[i]];
        }
        dp1[i + 1][j] |= dp1[i][j];
    }

    if (!dp1[n][k]) {
        OutPutInteger(-1);
        return 0;
    }
    reverse(all(a));

    rep(i, n)rep2(j, 0, k) {
        if (j - a[i] >= 0) {
            dp2[i + 1][j] |= dp2[i][j - a[i]];
        }
        dp2[i + 1][j] |= dp2[i][j];
    }

    int res = 0;
    rep(i, n) {
        bool exist = true;
        rep2(j, 0, k) {
            if (dp1[i][j] and dp2[n - 1 - i][k - j]) exist = false;
        }

        if (exist) res++;
    }

    if(res == 0) OutPutInteger(-1);
    else OutPutInteger(res);
 }