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;
using ll = long long;
using ull = unsigned long long;
static const ll INF = 1e+15;
int main() {
  int N, x;
  cin >> N >> x;
  vector<ll> a(N);
  for (int i = 0; i < N; ++i) {
    cin >> a[i];
  }
  ll ans = 0;
  ll max_move_cost = 0;
  for (int i = 0; i < N; ++i) {
    ll used_cost = 0;
    ll min_cost = INF;
    ll min_move_cost = INF;
    for (int j = 0; j < N; ++j) {
      ll move_count;
      if (j < i) {
        move_count = i - j;
      } else if (j > i) {
        move_count = N - (j - i);
      } else {
        move_count = 0;
      }
      ll move_cost = move_count * x;
      if (a[j] + move_cost < min_cost) {
        min_cost = a[j] + move_cost;
        used_cost = a[j];
        min_move_cost = move_cost;
      }
    }
    ans += used_cost;
    max_move_cost = max(max_move_cost, min_move_cost);
  }
  ans += max_move_cost;
  cout << ans << endl;
  return 0;
}
