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 <bits/stdc++.h>
using namespace std;
using ll = long long;
static const ll INF = 1e18;
vector<ll> d;
void dijkstra(const vector<vector<pair<int, ll>>> &G, int s) {
  priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>>
      pq;
  d.resize(G.size(), INF);
  d[s] = 0;
  pq.push(make_pair(0, s));
  while (!pq.empty()) {
    pair<ll, int> p = pq.top();
    pq.pop();
    int v = p.second;
    if (d[v] < p.first) continue;
    for (int i = 0; i < G[v].size(); i++) {
      pair<int, ll> p2 = G[v][i];
      if (d[p2.first] > d[v] + p2.second) {
        d[p2.first] = d[v] + p2.second;
        pq.push(make_pair(d[p2.first], p2.first));
      }
    }
  }
  return;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<ll> a(n);
  for (int i = 0; i < n; i++) cin >> a.at(i);
  vector<vector<pair<int, ll>>> G(n + 1);
  for (int i = 0; i < n; i++) G[0].emplace_back(make_pair(i + 1, a.at(i)));
  for (int i = 0; i < n; i++) {
    if (i + 1 < n)
      G[i + 1].emplace_back(make_pair(i + 2, x));
    else
      G[i + 1].emplace_back(make_pair(1, x));
  }
  dijkstra(G, 0);
  ll lasta = 1e18;
  vector<ll> xmax(n, 1e18), cost(n, 1e18);
  for (int i = 1; i <= n; i++) {
    if (d[i] == a.at(i - 1)) {
      lasta = a.at(i - 1);
      cost[i - 1] = a.at(i - 1);
      xmax[i - 1] = 0;
      continue;
    }
    if (lasta == 1e18) continue;
    cost[i - 1] = lasta;
    if (i - 2 >= 0) xmax[i - 1] = xmax[i - 2] + 1;
  }
  for (int i = 1; i <= n; i++) {
    if (cost[i - 1] < 1e18) break;
    cost[i - 1] = lasta;
    if (i - 2 >= 0)
      xmax[i - 1] = xmax[i - 2] + 1;
    else
      xmax[i - 1] = xmax[n - 1] + 1;
  }
  ll ans = 0, xx = 0;
  for (int i = 0; i < n; i++) {
    ans += cost[i];
    xx = max(xx, xmax[i]);
  }
  ans += xx * x;
  cout << ans << '\n';
}
