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 lint = long long;
int main() {
  int N;
  lint x;
  cin >> N >> x;
  vector<lint> a(N);
  map<lint, int> place;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    place[a[i]] = i;
  }
  for (auto it = place.begin(); it != place.end(); it++) {
    int i = it->second;
    vector<lint> profit(N);
    profit[0] = 0;
    for (int j = 1; j < N; j++) {
      int pos = (i + j) % N;
      profit[j] = profit[j - 1] + a[i] + x - a[pos];
    }
    int rep_max =
        distance(profit.begin(), min_element(profit.begin(), profit.end()));
    for (int j = 1; j <= rep_max; j++) {
      int pos = (i + j) % N;
      a[pos] = a[i] + x;
    }
  }
  lint ans = 0;
  for (int i = 0; i < N; i++) {
    ans += a[i];
  }
  cout << ans << endl;
  return 0;
}
