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;
long long ans[2005][2005], a[2005], pmin[2005][2005], idx[2005][2005];
long long min(long long a, long long b) { return a < b ? a : b; }
int main() {
  int n;
  long long x;
  cin >> n >> x;
  for (int i = (1); i <= (n); i++) scanf("%lld", &a[i]);
  for (int i = (1); i <= (n); i++) {
    for (int j = (1); j <= (n); j++)
      if (i != j) {
        if (j > i)
          ans[i][n - j + i] = a[j] + x * (n - j + i);
        else
          ans[i][i - j] = a[j] + x * (i - j);
      }
  }
  for (int i = (1); i <= (n); i++) {
    pmin[i][0] = a[i];
    idx[i][0] = 0;
    for (int j = (1); j <= (n - 1); j++)
      if (pmin[i][j - 1] > ans[i][j])
        pmin[i][j] = ans[i][j], idx[i][j] = j;
      else
        pmin[i][j] = pmin[i][j - 1], idx[i][j] = idx[i][j - 1];
  }
  long long res = 0x7fffffff;
  for (int i = (0); i <= (n - 1); i++) {
    long long tmp = 0, t = n;
    for (int j = (1); j <= (n); j++) {
      tmp += pmin[j][i];
      t = min(t, idx[j][i]);
    }
    res -= t * x;
    res = min(res, tmp);
  }
  printf("%lld\n", res);
}
