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;
const int INF = 1e9;
int main() {
  int n, x;
  cin >> n >> x;
  vector<int> a;
  for (int i = 0; i < (n); i++) {
    int b;
    cin >> b;
    a.push_back(b);
  }
  int d[n];
  int p = 0;
  int now = 0;
  int res = 0;
  d[0] = a[0];
  for (int i = 1; i < n; i++) {
    d[i] = min(d[i - 1] + x + a[0], INF);
  }
  for (int i = 1; i < n; i++) {
    d[i] = min(d[i], a[i]);
    int m = 1;
    for (int j = (i + 1) % n; j != i; j = (j + 1) % n) {
      int u;
      if ((a[i] * (m + 1) + x * m) >= INF)
        u = INF;
      else
        u = a[i] * (m + 1) + x * m;
      d[j] = min(d[j], u);
      m++;
    }
  }
  res = 0;
  int e = -1;
  for (int i = 0; i < (n); i++) {
    if (d[i] == a[i]) {
      int j = (i + 1) % n;
      while (d[j] != a[j]) j = (j + 1) % n;
      if (j == 0)
        j = n - 1;
      else
        j--;
      int r;
      if (j >= i)
        r = j - i;
      else
        r = (n + j) - i;
      e = max(r, e);
      res += a[i] * (r + 1);
    }
  }
  cout << res + x * e << endl;
  return 0;
}
