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;
int main() {
  long long int n, x;
  cin >> n >> x;
  vector<vector<long long int> > wf(n + 1, vector<long long int>(n + 1, 1e9));
  for (int i = 1; i < n + 1; i++) {
    int a;
    cin >> a;
    wf[0][i] = a;
  }
  for (int i = 1; i < n; i++) {
    wf[i][i + 1] = x;
  }
  wf[n][1] = x;
  vector<vector<long long int> > fwf(n + 1, vector<long long int>(n + 1));
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < n + 1; j++) {
      fwf[i][j] = j;
    }
  }
  for (int k = 0; k <= n; k++) {
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        if (wf[i][j] > wf[i][k] + wf[k][j]) {
          wf[i][j] = wf[i][k] + wf[k][j];
          fwf[i][j] = fwf[i][k];
        }
      }
    }
  }
  long long int ans = 0;
  long long int maxr = 0;
  for (int i = 1; i < n + 1; i++) {
    ans += wf[0][fwf[0][i]];
    maxr = max(maxr, wf[0][i] - wf[0][fwf[0][i]]);
  }
  cout << ans + maxr << endl;
}
