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 argc, char const *argv[]) {
  long long int N, x, ans = 100000000000000, tans = 0;
  std::cin >> N >> x;
  std::vector<long long int> a(N);
  for (int i = 0; i < N; i++) std::cin >> a[i];
  for (int j = 0; j < N; j++) {
    std::vector<std::vector<long long int> > min_cost =
        vector<vector<long long int> >(N, vector<long long int>(2, 0));
    min_cost[0][0] = a[0];
    for (int i = 1; i < N; i++) {
      if (min_cost[0][0] > a[i] + (N - i) * x && N - i <= j) {
        min_cost[0][0] = a[i];
        min_cost[0][1] = N - i;
      }
    }
    tans = 0;
    for (int i = 1; i < N; i++) {
      if (a[i] <= min_cost[i - 1][0]) {
        min_cost[i][0] = a[i];
      } else if (min_cost[i - 1][1] < j) {
        min_cost[i][1] = min_cost[i - 1][1] + 1;
        min_cost[i][0] = min_cost[i - 1][0];
      } else {
        min_cost[i][0] = a[i];
      }
    }
    for (int i = 0; i < N; i++) {
      tans += min_cost[i][0];
    }
    ans = min(ans, tans + j * x);
  }
  std::cout << ans << '\n';
  return 0;
}
