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(void) {
  cin.sync_with_stdio(false);
  int N, x;
  cin >> N >> x;
  vector<long long> As(N);
  for (int n = 0; n < (int)(N); ++n) cin >> As[n];
  vector<long long> Bs;
  auto itr = min_element((As).begin(), (As).end());
  auto mn_idx = distance(begin(As), itr);
  for (int i = (mn_idx); i < (int)(N); ++i) {
    Bs.push_back(As[i]);
  }
  for (int i = 0; i < (int)(mn_idx); ++i) {
    Bs.push_back(As[i]);
  }
  As = Bs;
  long long longest_streak = 0;
  long long ans = As[0];
  for (int n = (1); n < (int)(N); ++n) {
    long long cost = As[n];
    for (int n2 = n - 1; n2 >= 0; --n2) {
      long long diff = n - n2;
      long long cost2 = diff * x;
      if (cost2 < cost) {
        longest_streak = max(longest_streak, diff);
        cost = As[n2];
      }
    }
    ans += cost;
  }
  ans += longest_streak * x;
  cout << ans << endl;
  return 0;
}
