#include <iostream>
#include <vector>
#include <algorithm>
#include <array>
#include <set>
#include <map>
#include <queue>
#include <climits>
#include <tuple>
#include <unordered_set>
#include <unordered_map>
#include <functional>
#include <cstdio>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < (n); ++(i))
#define repeat_reverse(i,n) for (int i = (n)-1; (i) >= 0; --(i))
#define repeat_from_reverse(i,m,n) for (int i = (n)-1; (i) >= (m); --(i))
#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
typedef long long ll;
using namespace std;
int main() {
    int n; ll v; cin >> n >> v;
    vector<ll> c(n); repeat (i,n) cin >> c[i];
    if (v <= n) {
        cout << accumulate(c.begin(), c.end(), 0ll) << endl;
        return 0;
    }
    v -= n; // the restriction, that all fruits must be used, is fulfilled now.
    vector<ll> acc(n+1); repeat (i,n) acc[i+1] = acc[i] + c[i];
    const int l = n*n*n;
    vector<ll> dp(l+1, LLONG_MAX);
    dp[0] = 0;
    repeat (i,l) {
        repeat (j,min(i+1,n)) {
            dp[i+1] = min(dp[i+1], dp[i-j] + acc[j+1]);
        }
    }
    if (v <= l) {
        cout << acc[n] + dp[v] << endl;
        return 0;
    }
    vector<int> eff(l); repeat (i,l) eff[i] = i+1;
    sort(eff.begin(), eff.end(), [&](int a, int b) {
        return dp[a] * b < dp[b] * a; // sort by the efficiency
    });
    int e = eff.front();
    cout << acc[n] + (v / e) * dp[e] + dp[v % e] << endl;
    return 0;
}