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>
const int INF = 2000000000;
using namespace std;
const int MAX_N = 1 << 17;
int n, dat[2 * MAX_N - 1];
void init(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < (2 * n - 1); i++) dat[i] = INT_MAX;
}
void update(int k, int a) {
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
  }
}
int query(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) return INT_MAX;
  if (a <= l && r <= b)
    return dat[k];
  else {
    int vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    int vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
}
int main() {
  int N;
  long long x;
  cin >> N >> x;
  init(N);
  long long a[2005];
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
    update(i, a[i]);
  }
  long long ans = INF;
  for (int i = (0); i < (N); i++) {
    long long cost = 0;
    for (int j = 0; j < (N); j++) {
      if (j - i < 0) {
        cost += min(query(0, j + 1, 0, 0, n), query(N - i + j, N, 0, 0, n));
      } else {
        cost += query(j - i, j + 1, 0, 0, n);
      }
    }
    cost += x * i;
    ans = min(ans, cost);
  }
  cout << ans << endl;
}
