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 N, x;
int a[2020];
vector<int> kouho;
int len;
long ans = 0;
int get_a(int index) {
  int i = 0;
  if (1 <= index && index <= N)
    i = index;
  else if (index < 1)
    i = N + index;
  else if (index > N)
    i = index - N;
  return a[i];
}
int main() {
  cin >> N >> x;
  for (int i = 1; i < N + 1; i++) cin >> a[i];
  len = (N - 1) / 2;
  if (N % 2 == 0) {
    for (int i = 1; i < N + 1; i++) {
      kouho.clear();
      kouho.push_back(get_a(i));
      for (int j = 1; j <= len; j++) kouho.push_back(get_a(i + j) + x * j);
      for (int j = 1; j <= len + 1; j++) kouho.push_back(get_a(i - j) + x * j);
      sort(kouho.begin(), kouho.end());
      reverse(kouho.begin(), kouho.end());
      ans += kouho.back();
    }
  }
  if (N % 2 == 1) {
    for (int i = 1; i < N + 1; i++) {
      kouho.clear();
      kouho.push_back(get_a(i));
      for (int j = 1; j <= len; j++) kouho.push_back(get_a(i + j) + x * j);
      for (int j = 1; j <= len; j++) kouho.push_back(get_a(i - j) + x * j);
      sort(kouho.begin(), kouho.end());
      reverse(kouho.begin(), kouho.end());
      ans += kouho.back();
    }
  }
  cout << ans << endl;
  return 0;
}
