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;

struct edge {
    ll cost;
    int to;
};

bool operator<(const edge &lhs, const edge &rhs) {
    return lhs.cost < rhs.cost;
}
bool operator>(const edge &lhs, const edge &rhs) {
    return rhs.cost < lhs.cost;
}

struct queue_node {
    ll cost;
    int from;
};

bool operator<(const queue_node &lhs, const queue_node &rhs) {
    return lhs.cost < rhs.cost;
}
bool operator>(const queue_node &lhs, const queue_node &rhs) {
    return rhs.cost < lhs.cost;
}

struct result_node {
    ll cost;
    int step;
};

bool operator<(const result_node &lhs, const result_node &rhs) {
    return lhs.cost < rhs.cost;
}
bool operator>(const result_node &lhs, const result_node &rhs) {
    return rhs.cost < lhs.cost;
}

vector<result_node> dijkstra(vector<vector<edge>> const &graph, int start,
                             int step_limit) {
    vector<result_node> result(graph.size(), {LNF, 0});
    priority_queue<queue_node, vector<queue_node>, greater<queue_node>> que;
    result[start] = {0, 0};
    que.push({0, start});

    while (!que.empty()) {
        auto c = que.top();
        que.pop();
        // if (c.cost > result[c.from].cost) continue;
        for (auto e : graph[c.from]) {
            if (result[e.to].cost > result[c.from].cost + e.cost &&
                result[c.from].step < step_limit) {
                result[e.to].cost = result[c.from].cost + e.cost;
                result[e.to].step = result[c.from].step + 1;
                que.push({result[e.to].cost, e.to});
            }
        }
    }

    return result;
}

int main() {
    int N;
    ll x;
    cin >> N >> x;
    vector<vector<edge>> graph(N + 1);
    for (int i = 0; i < N; i++) {
        ll cost;
        cin >> cost;
        graph[0].push_back({cost, i + 1});
        graph[(i + (N - 1)) % N + 1].push_back({0, i + 1});
    }

    ll cand = LNF;
    for (int limit = 1; limit <= N; limit++) {
        auto result    = dijkstra(graph, 0, limit);
        auto max_steps = accumulate(
            result.begin(), result.end(), 0,
            [](auto const &s, auto const &a) { return max(s, a.step); });
        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.cost; });
        cand = min(cand, sum_summon + x * max_steps);
    }
    cout << cand << endl;

    return 0;
}
