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 MOD = static_cast<int>(1e9) + 7;
using namespace std::rel_ops;
inline long long bit(int x) { return 1LL << x; }
template <class T>
inline int size(const T& a) {
  return (int)a.size();
}
template <class T>
inline bool chmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
inline int clamp(T v, T lo, T hi) {
  return (v > hi) - (v < lo);
}
template <class T>
inline bool chmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
inline pair<vector<T>, char> operator*(const vector<T>& v, const char& c) {
  return make_pair(v, c);
}
template <class T>
inline istream& operator>>(istream& is, vector<T>& v) {
  for (auto& x : v) is >> x;
  return is;
}
template <class T>
inline ostream& operator<<(ostream& os, const pair<vector<T>, char>& v) {
  bool t = 0;
  for (const T& x : v.first) {
    if (t) os << v.second;
    os << x;
    t = 1;
  }
  return os;
}
template <class T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  return os << v * ' ';
}
struct before_main {
  before_main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
    ;
  };
} __before_main;
int main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  cin >> a;
  long long ans = accumulate(begin(a), end(a), 0LL);
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < n; j++) {
      a[j] = min(a[j], a[(j - i + n) % n]);
    }
    ans = min(ans, accumulate(begin(a), end(a), i * x));
  }
  cout << ans << endl;
}
