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;
long long N, x;
long long A[2001];
long long C[2001][2001];
pair<long long, long long> D[2001][2001];
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    C[i][0] = A[i];
    D[i][0] = pair<long long, long long>(i, 0);
  }
  for (int j = 1; j < N; j++) {
    for (int i = 0; i < N; i++) {
      if (C[i][j - 1] > C[(i + N - j) % N][0] + j * x) {
        C[i][j] = C[(i + N - j) % N][0] + j * x;
        D[i][j] = pair<long long, long long>((i + N - j) % N, j);
      } else {
        C[i][j] = C[i][j - 1];
        D[i][j] = D[i][j - 1];
      }
    }
  }
  long long ans = 2e18;
  for (int j = 0; j < N; j++) {
    long long now = 0;
    long long magic = 0;
    for (int i = 0; i < N; i++) {
      now += A[D[i][j].first];
      magic = max(magic, D[i][j].second);
    }
    ans = min(now + magic * x, ans);
  }
  cout << ans << endl;
}
