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() {
  int n, x;
  cin >> n >> x;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long ls[n][n];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ls[i][j] = ((j <= i) ? a[j] + (i - j) * x : a[j] + (n - (j - i)) * x);
    }
  }
  int best[n], num[n];
  pair<long long, pair<int, int> > res[n];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j == 0) {
        best[i] = ls[i][j];
        num[i] = 0;
      } else {
        if (best[i] > ls[i][j]) {
          best[i] = ls[i][j];
          num[i] = j;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    res[i] =
        pair<long long, pair<int, int> >(i - num[i], pair<int, int>(num[i], i));
    if (res[i].first < 0) res[i].first += n;
  }
  sort(res, res + n);
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1)
      ans += ls[res[i].second.second][res[i].second.first];
    else
      ans += (ls[res[i].second.second][res[i].second.first] - res[i].first * x);
  }
  cout << ans << endl;
  return 0;
}
