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 int n, a, b, least, mg, sum, lim;
vector<long long int> po;
int main() {
  cin >> n >> mg;
  po.push_back(0);
  for (int i = 0; i < n; i++) {
    cin >> a;
    po.push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    lim = po[i] / mg;
    least = po[i];
    for (int j = 1; j <= lim; j++) {
      if (j == n + 1) break;
      if (i + j <= n) {
        if (po[i + j] + j * mg < least) least = po[i + j] + mg * j;
      } else {
        b = (i + j) % n;
        if (b == 0) b = 1;
        if (po[b] + j * mg < least && b <= n) least = po[b] + j * mg;
      }
      if (i - j > 0) {
        if (po[i - j] + mg * j < least) least = po[i - j] + mg * j;
      } else {
        b = n + 1 - (i + j) % n;
        if (b == 0) b = 1;
        if (po[b] + mg * j < least) least = po[b] + mg * j;
      }
    }
    sum += least;
  }
  cout << sum << endl;
  return 0;
}
