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;
  std::cin >> n >> x;
  std::vector<int> a;
  for (int i = 0; i < n; i++) {
    int temp;
    std::cin >> temp;
    a.push_back(temp);
  }
  std::vector<std::vector<long long int>> small;
  std::vector<long long int> smallf;
  smallf.push_back(a[0]);
  for (int i = 1; i < n; i++) {
    smallf.push_back(a[i]);
  }
  small.push_back(smallf);
  for (int j = 1; j < n; j++) {
    std::vector<long long int> smallt;
    if (small[j - 1][0] >= x + small[j - 1][n - 1]) {
      smallt.push_back(x + small[j - 1][n - 1]);
    } else {
      smallt.push_back(small[j - 1][0]);
    }
    for (int i = 1; i < n; i++) {
      if (small[j - 1][i] >= x + small[j - 1][i - 1]) {
        smallt.push_back(x + small[j - 1][i - 1]);
      } else {
        smallt.push_back(small[j - 1][i]);
      }
    }
    small.push_back(smallt);
  }
  long long int sum = 0;
  int flag = 0;
  int count = 0;
  int fsum = 0;
  for (int i = 0; i < n; i++) {
    count = 0;
    int tf = 0;
    long long int min = small[0][i];
    for (int j = 1; j < n; j++) {
      count++;
      if (small[j][i] < min) {
        if (flag < count) {
          flag = count;
        }
        tf = count;
        min = small[j][i];
      }
    }
    fsum += tf;
    sum += min;
  }
  std::cout << sum + flag * x - fsum * x << std::endl;
  return 0;
}
