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;
struct input_returnner {
  int N;
  input_returnner(int N_ = 0) : N(N_) {}
  template <typename T>
  operator vector<T>() const {
    vector<T> res(N);
    for (auto &a : res) cin >> a;
    return std::move(res);
  }
  template <typename T>
  operator T() const {
    T res;
    cin >> res;
    return res;
  }
};
template <typename T>
input_returnner in() {
  return in<T>();
}
input_returnner in() { return input_returnner(); }
input_returnner in(int N) { return std::move(input_returnner(N)); }
signed main(void) {
  cin.tie(0), ios::sync_with_stdio(false), cout << setprecision(15);
  int N, X;
  cin >> N >> X;
  vector<int> A = in(N);
  vector<int> check;
  for (int i = 0; i < (N); i++) {
    bool f = true;
    for (int j = (1); j < (N); j++) {
      if (A[i] >= A[(i - j + N) % N] + j * X) f = false;
    }
    if (f) check.push_back(i);
  }
  vector<bool> b(N);
  int res = 0;
  for (int i = 0; i < (N); i++) {
    for (auto &c : check) {
      if (not b[c]) {
        b[c] = true;
        res += A[c];
      }
    }
    vector<bool> s(N);
    bool flag = true;
    for (int j = 0; j < (N); j++) {
      s[(j + 1) % N] = b[j];
      if (not b[j]) flag = false;
    }
    if (flag) break;
    res += X;
    b = s;
  }
  cout << res << endl;
  return 0;
}
