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 slime {
  long long a;
  int pos;
  bool operator<(const slime& right) const { return a < right.a; }
};
int main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> a(n);
  vector<struct slime> slimes(n);
  vector<bool> ok(2 * n, false);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    slimes[i] = {a[i], i};
  }
  sort(slimes.begin(), slimes.end());
  long long cost;
  long long ans = 0;
  long long answer = 10000000000000;
  for (int kaisu = 0; kaisu < n; kaisu++) {
    ans = 0;
    ans += kaisu * x;
    for (int i = 0; i < 2 * n; i++) {
      ok[i] = false;
    }
    for (int i = 0; i < n; i++) {
      cost = 10000000000000;
      for (int j = slimes[i].pos + n - 1; j > slimes[i].pos + n - 1 - kaisu;
           j--) {
        if (ok[j]) {
          cost = ((cost) < (a[j]) ? (cost) : (a[j]));
        }
      }
      if (cost < slimes[i].a) {
        ans += cost;
      } else {
        ok[slimes[i].pos] = ok[slimes[i].pos + n] = true;
        ans += slimes[i].a;
      }
    }
    answer = ((answer) < (ans) ? (answer) : (ans));
  }
  cout << answer << endl;
  return 0;
}
