#include <bits/stdc++.h>

#define show(x) cout << #x << " = " << x << endl

using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "sz=" << v.size() << "\n[";
    for (const auto& p : v) {
        os << p << ",";
    }
    os << "]\n";
    return os;
}

template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p)
{
    os << "(" << p.first << "," << p.second
       << ")";
    return os;
}


constexpr ll MOD = 1e9 + 7;

template <typename T>
constexpr T INF = numeric_limits<T>::max() / 100;

int main()
{
    int N;
    ll V;
    cin >> N >> V;
    vector<ll> cost(N);
    ll suma = 0;
    for (int i = 0; i < N; i++) {
        cin >> cost[i];
        suma += cost[i];
    }
    if (V <= N) {
        cout << suma << endl;
        return 0;
    }
    V -= N;
    vector<ll> sum(N);
    vector<pair<long double, int>> newcost(N);
    for (int i = 0; i < N; i++) {
        if (i == 0) {
            sum[i] = cost[i];
        } else {
            sum[i] = sum[i - 1] + cost[i];
        }
        newcost[i] = make_pair((long double)sum[i] / (i + 1), i);
    }
    sort(newcost.begin(), newcost.end());

    constexpr ll REST = 50000;
    const ll res = min(REST, V);
    const int ind = newcost[0].second;
    const ll vol = (V - res) / (ind + 1);
    V -= vol * (ind + 1);
    suma += vol * sum[ind];


    vector<vector<ll>> dp(N, vector<ll>(V + 1, INF<ll>));  //i番目まででLリットル作るときの最小コスト
    for (ll i = 0; i <= V; i++) {
        dp[0][i] = sum[0] * i;
    }

    for (ll i = 1; i < N; i++) {
        for (ll l = 0; l <= V; l++) {
            dp[i][l] = dp[i - 1][l];
        }
        for (ll l = 0; l <= V; l++) {
            if (l >= (i + 1)) {
                dp[i][l] = min(dp[i][l], dp[i][l - (i + 1)] + sum[i]);
            }
        }
    }
    cout << suma + dp[N - 1][V] << endl;

    return 0;
}