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;
long long N, x;
long long a[2000];
long long magic[2001];
pair<long long, long long> d[2010];
signed main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    d[i].first = a[i];
    d[i].second = 0;
  }
  bool change = true;
  while (change) {
    change = false;
    for (int i = 0; i < N; i++) {
      if (i == 0) {
        if (d[i].first > d[N - 1].first + x) {
          d[i].first = d[N - 1].first + x;
          d[i].second += d[N - 1].second + 1;
          change = true;
        }
      } else {
        if (d[i].first > d[i - 1].first + x) {
          d[i].first = d[i - 1].first + x;
          d[i].second += d[i - 1].second + 1;
          change = true;
        }
      }
    }
  }
  for (int i = 0; i < 2001; i++) magic[i] = 0;
  long long ans = 0;
  long long magic_max_cnt = 0;
  for (int i = 0; i < N; i++) {
    ans += d[i].first - x * d[i].second;
    magic_max_cnt = max(magic_max_cnt, d[i].second);
    magic[d[i].second]++;
  }
  ans += x * magic_max_cnt;
  cout << ans << endl;
}
