#include <bits/stdc++.h>
using namespace std;
 
const int N = 101;
const int mx = 1e6+1;
int n, S, a[N], dp[mx];
// Gọi dp[i] là số giờ ít nhất để tạo thành số dinh dưỡng i
// Bài toán quy về dạng dijsktra:  Từ đỉnh u có thể đi đến các đỉnh  v = u + a[i] (v <= S) với w = dp[u] + 1 + i
 
int main(){
//  icebear hehehhe
    // freopen("DOCTOR.inp", "r", stdin);
    // freopen("DOCTOR.out", "w", stdout);
    cin.tie(0) -> sync_with_stdio(0);
    cin >> n >> S;
    for(int i=1; i<=n; ++i) cin >> a[i];
    priority_queue<pair<int, int>> q;
    memset(dp, 0x3f, sizeof dp);
    q.push(make_pair(-1, 0));
    dp[0] = -1;
    while(!q.empty()){
        int du = q.top().first;
        int u = q.top().second;
        q.pop();
 
        if (du != dp[u]) continue;
        if (u == S){
            cout << dp[u];
            exit(0);
        }
 
        for(int i=1; i<=n; ++i){
            u += a[i];
            if (u > S) break;
            if (dp[u] > du + i + 1)
                q.push(make_pair(dp[u] = du + i + 1, u));
        }
    }
    cout << -1;
    return 0;
}
 

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