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;
int dy[4] = {-1, 1, 0, 0};
int dx[4] = {0, 0, 1, -1};
int N, x;
int a[10];
class rminq {
 public:
  int n;
  vector<int> dat;
  rminq(int n_);
  void update(int k, int a);
  int query(int a, int b, int k, int l, int r);
  int query(int a, int b);
};
rminq::rminq(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  dat.resize(2 * n - 1, 1000000007LL);
}
void rminq::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 rminq::query(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) return 1000000007LL;
  if (a <= l && r <= b) return dat[k];
  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 rminq::query(int a, int b) { return query(a, b, 0, 0, n); }
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> x;
  for (long long i = 0; i < (long long)(N); i++) cin >> a[i];
  rminq m(N);
  for (long long i = 0; i < (long long)(N); i++) {
    m.update(i, a[i]);
  }
  int ans = 1000000007LL;
  for (long long k = 0; k < (long long)(N); k++) {
    int tmp_ans = k * x;
    for (long long right = 0; right < (long long)(N); right++) {
      int left = (N + right - k) % N;
      if (left > right) {
        tmp_ans += min(m.query(left, N), m.query(0, right + 1));
      } else {
        tmp_ans += m.query(left, right + 1);
      }
    }
    ans = min(tmp_ans, ans);
  }
  cout << ans << endl;
  return 0;
}
