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;
const double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342;
const int mod = 1000000007;
int main() {
  long long n, x, a[2222] = {}, m[2222] = {}, m2[2222] = {};
  cin >> n >> x;
  for (int i = (0); i < (n); ++i) {
    cin >> a[i];
    m[i] = a[i];
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      long long nm = a[i] + j * x;
      if (nm < m[(i + j) % n]) {
        m[(i + j) % n] = nm;
        m2[(i + j) % n] = j;
      }
    }
  }
  long long mx = 0;
  for (int i = (0); i < (n); ++i) mx = max(m2[i], mx);
  long long ans = 0LL;
  for (int i = (0); i < (n); ++i) ans += m[i] - x * m2[i];
  ans += mx * x;
  cout << ans;
  return 0;
}
