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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
const long long INF = 1e14;
int main() {
  int n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = 0; i < (int)(n); i++) {
    cin >> a[i];
  }
  vector<pair<long long, int>> min_cost(n, make_pair(INF, 0));
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(n); j++) {
      if (i == j) {
        if (min_cost[i].first > a[i]) {
          min_cost[i].first = a[i];
          min_cost[i].second = 0;
        }
      } else if (i > j) {
        if (min_cost[i].first >= a[j] + (i - j) * x) {
          min_cost[i].first = a[j] + (i - j) * x;
          min_cost[i].second = (i - j);
        }
      } else if (i < j) {
        if (min_cost[i].first >= a[j] + (j + (n - i)) * x) {
          min_cost[i].first = a[j] + (i + (n - j)) * x;
          min_cost[i].second = (j + (n - i));
        }
      }
    }
  int slide_cnt = 0, slide_sum = 0;
  long long ans = 0;
  for (int i = 0; i < (int)(n); i++) {
    ans += min_cost[i].first;
    slide_cnt = max(min_cost[i].second, slide_cnt);
    slide_sum += min_cost[i].second;
  }
  cout << ans - (slide_sum - slide_cnt) * x << endl;
  return 0;
}
