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 long long linf = 1e18;
const double eps = 1e-12;
const double pi = acos(-1);
template <typename T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (auto&& x : vec) is >> x;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  for (long long i = (0), __last_i = (vec.size()); i < __last_i; i++) {
    if (i) os << " ";
    os << vec[i];
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<vector<T> >& vec) {
  for (long long i = (0), __last_i = (vec.size()); i < __last_i; i++) {
    if (i) os << endl;
    os << vec[i];
  }
  return os;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  cin >> a;
  vector<long long> b = a;
  for (long long i = (0), __last_i = (n); i < __last_i; i++)
    for (long long j = (0), __last_j = (n); j < __last_j; j++) {
      long long d = i >= j ? i - j : n + i - j - 1;
      b[i] = min(b[i], a[j] + x * d);
    }
  long long ans = 0;
  for (long long i = (0), __last_i = (n); i < __last_i; i++) ans += b[i];
  cout << ans << endl;
}
