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 MOD = 1e9 + 7;
typedef struct {
  long long to;
  long long cost;
} edge;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long N, x;
  long long a[2010];
  cin >> N >> x;
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
  }
  sort(a, a + N);
  long long min_val[2010] = {};
  long long min_pos[2010] = {};
  for (int i = 0; i < (N); i++) {
    min_val[i] = (1e18);
    min_pos[i] = -1;
  }
  for (int i = 0; i < (N); i++) {
    for (int j = 0; j < (N); j++) {
      if (i == j) {
        if (min_val[i] > a[i]) {
          min_val[i] = a[i];
          min_pos[i] = j;
        }
      } else if (j < i) {
        if (min_val[i] > a[j] + x * (i - j)) {
          min_val[i] = a[j] + x * (i - j);
          min_pos[i] = j;
        }
      } else {
        if (min_val[i] > a[j] + x * (i + N - j)) {
          min_val[i] = a[j] + x * (i + N - j);
          min_pos[i] = j;
        }
      }
    }
  }
  long long diff_max = 0;
  long long diff_arr[100010] = {};
  for (int i = 0; i < (N); i++) {
    diff_arr[i - min_pos[i]]++;
  }
  long long ans = 0;
  ans += min_val[N - 1];
  for (int i = 0; i < (N - 1); i++) {
    ans += a[min_pos[i]];
  }
  cout << (ans) << endl;
  ;
}
