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;
constexpr double PI = 3.1415926535897932;
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, x;
  cin >> n >> x;
  vector<long long> a(n);
  using Pl = pair<long long, int>;
  vector<Pl> time(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    time[i] = pair<int, int>(a[i], i);
  }
  for (int i = 0; i < n; ++i) {
    long long cur = a[i] + x;
    for (int j = i + 1; j < n; ++j) {
      if (time[j].first >= cur) {
        if (time[j].first == cur) {
          int nowgap = j - i;
          int pregap;
          if (j >= time[j].second)
            pregap = j - time[j].second;
          else
            pregap = n - time[j].second + j;
          if (nowgap < pregap) {
            time[j] = Pl(cur, i);
          }
        } else
          time[j] = Pl(cur, i);
      }
    }
    for (int j = 0; j < i; ++j) {
      if (time[j].first >= cur) {
        if (time[j].first == cur) {
          int nowgap = n - i + j;
          int pregap;
          if (j >= time[j].second)
            pregap = j - time[j].second;
          else
            pregap = n - time[j].second + j;
          if (nowgap < pregap) {
            time[j] = Pl(cur, i);
          }
        } else
          time[j] = Pl(cur, i);
      }
    }
  }
  long long ans = 0;
  int maxgap = 0;
  for (int i = 0; i < n; ++i) {
    ans += time[time[i].second].first;
    int gap;
    if (i >= time[i].second)
      gap = i - time[i].second;
    else
      gap = n - time[i].second + i;
    maxgap = max(maxgap, gap);
  }
  ans += (long long)x * maxgap;
  cout << ans << endl;
}
