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 PT {
  signed long long x;
  signed long long y;
};
string to_s(unsigned long long x) { return to_string(x); }
string to_s(signed long long x) { return to_string(x); }
signed long long to_ll(const string& x) { return stoll(x); }
unsigned long long to_ull(const string& x) { return stoull(x); }
int main(int argc, char* argv[]) {
  cout << std::fixed << std::setprecision(10);
  for (;;) {
    int N, X;
    cin >> N >> X;
    if (cin.fail()) break;
    vector<int> A(N);
    for (int i = 0; i < N; i++) {
      cin >> A[i];
    }
    vector<signed long long> S(N, (1LL << 48));
    signed long long res = (1LL << 48);
    for (int i = 0; i < N; i++) {
      signed long long old = S[N - 1];
      for (int j = N - 1; j > 0; j--) {
        S[j] = S[j - 1];
      }
      S[0] = old;
      for (int j = 0; j < N; j++) {
        signed long long cost = (signed long long)i * X + A[j];
        if (cost < S[j]) {
          S[j] = cost;
        }
      }
      signed long long r = 0;
      for (int i = 0; i < N; i++) {
        r += S[i];
      }
      res = min(res, r);
    }
    cout << res << endl;
  }
  return 0;
}
