Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

* Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

* Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.




Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

* 2≤N≤2,000
* a_i are integers.
* 1≤a_i≤10^9
* x is an integer.
* 1≤x≤10^9

Input

The input is given from Standard Input in the following format:


N x
a_1 a_2 ... a_N


Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
#include <limits>
using ll = long long;
constexpr int INF = std::numeric_limits<int>::max() / 2;
constexpr ll LNF = std::numeric_limits<ll>::max() / 2;

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

vector<pair<ll, int>> dijkstra(vector<vector<pair<int, ll>>> const &graph,
                               int start) {
    vector<pair<ll, int>> result(graph.size(), make_pair(LNF, 0));
    priority_queue<pair<ll, int>, vector<pair<ll, int>>,
                   greater<pair<ll, int>>>
        que;
    result[start] = make_pair(0, 0);
    que.push(make_pair(0, start));

    ll cost;
    int from;
    while (!que.empty()) {
        tie(cost, from) = que.top();
        que.pop();
        if (cost > result[from].first) continue;
        for (auto to : graph[from]) {
            if (result[to.first].first > result[from].first + to.second) {
                result[to.first].first  = result[from].first + to.second;
                result[to.first].second = result[from].second + 1;
                que.push(make_pair(result[to.first].first, to.first));
            }
        }
    }

    return result;
}

int main() {
    int N;
    ll x;
    cin >> N >> x;
    vector<vector<pair<int, ll>>> graph(N + 1);
    for (int i = 0; i < N; i++) {
        ll cost;
        cin >> cost;
        graph[0].push_back(make_pair(i + 1, cost));
        graph[(i + (N - 1)) % N + 1].push_back(make_pair(i + 1, x));
    }
    auto result    = dijkstra(graph, 0);
    auto max_steps = accumulate(
        result.begin(), result.end(), 0,
        [](auto const &s, auto const &a) { return max(s, a.second); });
    max_steps--;
    cerr << max_steps << endl;
    auto sum_summon =
        accumulate(result.begin() + 1, result.end(), 0ll,
                   [&](auto const &s, auto const &a) {
                       return s + a.first - ((a.second - 1) * x);
                   });
    cout << sum_summon + x * max_steps << endl;

    return 0;
}
