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 std::cin;
const int INF = 1e+9;
const long long INFLL = 1e+17;
const long long MOD = 1e+9 + 7;
const int NMAX = 3000;
int N;
long long x;
long long A[2010];
class RMQ {
 public:
  long long dat[NMAX * 2 - 1];
  int n;
  RMQ(int n_);
  void update(int k, long long a);
  long long query(int a, int b, int k, int l, int r);
};
RMQ::RMQ(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < 2 * n - 1; i++) {
    dat[i] = INFLL;
  }
}
void RMQ::update(int k, long long a) {
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = std::min(dat[2 * k + 1], dat[2 * k + 2]);
  }
}
long long RMQ::query(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) return INFLL;
  if (a <= l && r <= b)
    return dat[k];
  else {
    long long v1 = query(a, b, 2 * k + 1, l, (l + r) / 2);
    long long v2 = query(a, b, 2 * k + 2, (l + r) / 2, r);
    return std::min(v1, v2);
  }
}
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) cin >> A[i];
  RMQ rmq(N);
  for (int i = 0; i < N; i++) {
    rmq.update(i, A[i]);
  }
  long long res = INFLL;
  for (long long num = 0; num < N; num++) {
    long long temp = x * num;
    for (int i = 0; i < N; i++) {
      if (i - num >= 0)
        temp += rmq.query(i - num, i + 1, 0, 0, N);
      else
        temp += std::min(rmq.query(0, i + 1, 0, 0, N),
                         rmq.query(N - (num - i), N, 0, 0, N));
    }
    res = std::min(res, temp);
  }
  std::cout << res << "\n";
  return 0;
}
