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;
int N;
long long x;
long long base = 1000000005LL;
int main() {
  cin >> N >> x;
  vector<long long> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];
  vector<vector<long long>> T(N, vector<long long>(N, 0LL));
  vector<vector<int>> S(N, vector<int>(N, 0));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (j == 0) {
        T[i][j] = A[i];
        S[i][j] = i;
      } else {
        if (T[i][j - 1] > A[(i - j + N) % N] + j * x) {
          T[i][j] = A[(i - j + N) % N] + j * x;
          S[i][j] = (i - j + N) % N;
        } else {
          T[i][j] = T[i][j - 1];
          S[i][j] = S[i][j - 1];
        }
      }
    }
  }
  long long min_s = 200000000000005LL;
  for (int i = 0; i < N; i++) {
    long long ss = 0LL;
    for (int j = 0; j < N; j++) {
      ss += A[S[j][i]];
    }
    ss += i * x;
    min_s = min(min_s, ss);
  }
  cout << min_s << endl;
}
