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;
struct edge {
  long long to;
  long long cost;
};
using P = pair<long long, long long>;
long long V;
vector<edge> G[2000 + 1];
long long d[2000 + 1];
long long shotai;
void dijkstra(long long s) {
  priority_queue<P, vector<P>, greater<P> > que;
  fill(d, d + V + 1, 1e+13);
  d[s] = 0;
  que.push(P(0, s));
  while (!que.empty()) {
    P p = que.top();
    que.pop();
    long long v = p.second;
    if (d[v] < p.first) continue;
    for (long long i = 0; i < G[v].size(); ++i) {
      edge e = G[v][i];
      if (d[e.to] > d[v] + e.cost) {
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
        if (e.to == 0) {
          shotai = v;
        }
      }
    }
  }
}
int main() {
  cin >> V;
  long long a[V + 1];
  long long x;
  long long i;
  cin >> x;
  for (i = 1; i <= V; i++) {
    cin >> a[i];
  }
  for (i = 1; i <= V; i++) {
    edge e1 = {0, a[i]};
    G[i].push_back(e1);
  }
  for (i = 2; i <= V; i++) {
    edge e2 = {i - 1, x};
    G[i].push_back(e2);
  }
  edge e3 = {V, x};
  G[1].push_back(e3);
  long long ans = 0;
  long long MAX_change = 0;
  for (i = 1; i <= V; i++) {
    dijkstra(i);
    ans += a[shotai];
    if (shotai <= i) {
      MAX_change = max(MAX_change, i - shotai);
    } else {
      MAX_change = max(MAX_change, V - (shotai - i));
    }
  }
  ans += MAX_change * x;
  cout << ans << endl;
  system("pause");
  return 0;
}
