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;
void Debug() { cout << '\n'; }
template <class FIRST, class... REST>
void Debug(FIRST arg, REST... rest) {
  cout << arg << " ";
  Debug(rest...);
}
template <class T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  out << "[";
  if (!v.empty()) {
    for (int i = int(0); i < int((int)(v).size() - 1); i++) out << v[i] << ", ";
    out << v.back();
  }
  out << "]";
  return out;
}
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& v) {
  out << "(" << v.first << ", " << v.second << ")";
  return out;
}
const int MAX_N = 200010;
const int MAX_V = 100010;
const double eps = 1e-6;
const long long mod = 1000000007;
const int inf = 1 << 29;
const long long linf = 1LL << 60;
const double PI = 3.14159265358979323846;
int N;
long long x;
int A[MAX_N];
int dp[2010][2010];
int loop(int a, int b) {
  if (dp[a][b] != -1) return dp[a][b];
  if (a == b) return A[a];
  return dp[a][b] = min(min(loop(a + 1, b), A[a]), min(loop(a, b - 1), A[b]));
}
void solve() {
  cin >> N >> x;
  for (int i = int(0); i < int(N); i++) {
    cin >> A[i];
    A[i + N] = A[i];
  }
  memset(dp, -1, sizeof(dp));
  long long res = linf;
  for (int i = int(0); i < int(N); i++) {
    long long tmp = 0;
    for (int j = int(N); j < int(2 * N); j++) {
      tmp += loop(j - i, j);
    }
    res = min(res, tmp + x * i);
  }
  cout << res << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed;
  cout.precision(20);
  srand((unsigned int)time(NULL));
  solve();
  return 0;
}
