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 int INF = 1e9 + 10;
const long long int LINF = 1LL * INF * INF;
const int MAXN = 2e5 + 10;
const int MAXM = 2e3 + 10;
priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;
long long int A[MAXM];
int main() {
  long long int n, m, k, a, b, x, y, q;
  int sum = 0;
  int cnt = 0;
  int mx = 0;
  long long int mn = INF;
  int cur = 0, idx = -1;
  int tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> A[i];
  long long int res = 0;
  for (int i = 0; i < n; i++) {
    mn = A[i];
    idx = 0;
    for (int j = 1; j < n; j++)
      if (A[(i + j) % n] + j * x < mn) {
        mn = A[(i + j) % n] + j * x;
        idx = j;
      }
    res += A[(i + idx) % n];
    mx = max(idx, mx);
  }
  res += mx * x;
  cout << res << "\n";
  return 0;
}
