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;
template <typename T>
void readvec(vector<T> &a);
void readindex(vector<long long> &a);
class RMQ_segment_tree {
  using data_type = pair<long long, long long>;

 public:
  vector<data_type> dat;
  long long N;
  data_type infinity;
  RMQ_segment_tree(long long n, data_type inf) {
    N = 1;
    while (n > N) N = N << 1;
    dat = vector<data_type>(2 * N - 1, inf);
    infinity = inf;
  }
  void update(long long k, data_type 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]);
    }
  }
  data_type minval(long long a, long long b) { return query(a, b, 0, 0, N); }
  data_type query(long long a, long long b, long long k, long long l,
                  long long r) {
    if (r <= a || b <= l) return infinity;
    if (a <= l && r <= b)
      return dat[k];
    else {
      data_type vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      data_type vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return min(vl, vr);
    }
  }
};
signed main() {
  long long N, x;
  cin >> N >> x;
  vector<long long> a(N);
  readvec(a);
  RMQ_segment_tree sgt(2 * N, pair<long long, long long>(1000000000000000, -1));
  for (long long i = (0); i < (N); i++)
    sgt.update(i, pair<long long, long long>(a[i] - x * i, i));
  for (long long i = (0); i < (N); i++)
    sgt.update(i + N, pair<long long, long long>(a[i] - x * (i + N), i));
  long long ans = 1000000000000000;
  for (long long m = (0); m < (N); m++) {
    long long tmp = x * m;
    for (long long i = (0); i < (N); i++) {
      long long i0 = sgt.minval(i + N - m, i + N + 1).second;
      tmp += a[i0];
    }
    ans = min(ans, tmp);
  }
  cout << ans;
  return 0;
}
template <typename T>
void readvec(vector<T> &a) {
  for (long long i = (0); i < (a.size()); i++) {
    cin >> a[i];
  }
}
void readindex(vector<long long> &a) {
  for (long long i = (0); i < (a.size()); i++) {
    cin >> a[i];
    a[i]--;
  }
}
