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 N, x;
  scanf("%lld%lld", &N, &x);
  long long a[2002];
  for (int i = (1); i <= (N); ++i) {
    scanf("%lld", &a[i]);
  }
  long long ans = 0;
  int cn = 0;
  int select = -1;
  for (int i = (1); i <= (N); ++i) {
    long long nans = 1000000000000;
    int ncn = 0;
    for (int j = (1); j <= (N); ++j) {
      if (i == j) {
        if (nans > a[i]) {
          ncn = 0;
          select = j;
          nans = a[i];
        }
      } else if (i < j) {
        if (nans > a[j] + ((N - (j - i)) * x)) {
          ncn = N - (j - i);
          select = j;
          nans = a[j] + ((N - (j - i)) * x);
        }
      } else {
        if (nans > a[j] + (i - j) * x) {
          ncn = i - j;
          select = j;
          nans = (a[j] + (i - j) * x);
        }
      }
    }
    cn = max(ncn, cn);
    ans += a[select];
  }
  ans += x * (long long)cn;
  printf("%lld\n", ans);
}
