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 eps = 1e-10;
const int MOD = 1000000007;
const int INF = 1000000000;
const long long int LINF = 1 << 30;
template <typename T>
void printv(vector<T> const& s) {
  for (int i = (0); i < (((int)(s).size())); i++) {
    cout << s[i] << " ";
  }
  cout << endl;
}
int main() {
  cin.tie(0);
  cout << setprecision(10);
  long long int n, x;
  cin >> n >> x;
  vector<long long int> a(n);
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
  }
  vector<vector<pair<long long int, int>>> g(n);
  for (int i = (0); i < (n); i++) {
    g[i].resize(n);
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (n); j++) {
      if (j < i) g[i][j] = pair<long long int, int>(a[j] + x * (i - j), i - j);
      if (j == i) g[i][j] = pair<long long int, int>(a[j], 0);
      if (j > i)
        g[i][j] = pair<long long int, int>(a[j] + x * (i + n - j), i + n - j);
    }
  }
  long long int ma = 0;
  vector<long long int> sec(n);
  for (int i = (0); i < (n); i++) {
    long long int mi = LINF;
    long long int tmp = 0;
    for (int j = (0); j < (n); j++) {
      if (mi > g[i][j].first) {
        mi = g[i][j].first;
        tmp = g[i][j].second;
        sec[i] = a[j];
      }
    }
    ma = max(ma, tmp);
  }
  long long int ans = x * ma;
  for (int i = (0); i < (n); i++) {
    ans += sec[i];
  }
  cout << ans << endl;
}
