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 long long INF = 12345678901LL;
long long cost[2000][2000] = {0};
int main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf(" %lld", &a[i]);
  for (int(i) = 0; (i) < (int)(n); ++(i))
    for (int(j) = 0; (j) < (int)(n); ++(j)) {
      cost[i][j] = a[j];
      if (j <= i)
        cost[i][j] += x * (i - j);
      else
        cost[i][j] += x * (n + i - j);
    }
  long long ans = 0;
  int max_mct = 0;
  for (int(i) = 0; (i) < (int)(n); ++(i)) {
    long long t = INF;
    int idx = 0, mct = n;
    for (int(j) = 0; (j) < (int)(n); ++(j)) {
      if (cost[i][j] < t) {
        t = cost[i][j];
        idx = j;
        if (j <= i)
          mct = i - j;
        else
          mct = n + i - j;
      } else if (cost[i][j] == t) {
        int t_mct;
        if (j <= i)
          t_mct = i - j;
        else
          t_mct = n + i - j;
        if (t_mct < mct) {
          idx = j;
          mct = t_mct;
        }
      }
    }
    ans += a[idx];
    max_mct = max(max_mct, mct);
  }
  ans += x * max_mct;
  cout << ans << endl;
  return 0;
}
