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;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
const long long int INF = 1e18;
int n;
long long int x;
long long int ar[2005];
long long int mn[4005][4005];
int main() {
  cin >> n >> x;
  for (int i = 0; i < (n); ++i) cin >> ar[i], ar[i + n] = ar[i];
  for (int i = 0; i < (n * 2); ++i) {
    mn[i][i] = INF;
    for (int j = (i); j < (n * 2); ++j) {
      mn[i][j + 1] = min(mn[i][j], ar[j - 1]);
    }
  }
  long long int res = INF;
  for (int i = 0; i < (n); ++i) {
    long long int tmp = 0;
    for (int j = 0; j < (n); ++j) {
      tmp += mn[j + n - i][j + n + 1];
    }
    tmp += x * i;
    chmin(res, tmp);
  }
  cout << res << endl;
  return 0;
}
