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, menor = LONG_MAX, answer = 0, val;
long long colors[2010];
bool ready[2010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  colors[0] = 0;
  cin >> n >> x;
  int inicial;
  for (int i = 0; i < n; ++i) {
    cin >> colors[i];
    if (colors[i] < menor) {
      menor = colors[i];
      inicial = i;
    }
  }
  answer += menor;
  int corrimientos = 0;
  menor += x;
  for (int i = inicial + 1, j = 1; j < n; ++i, ++j, i %= n) {
    if (colors[i] <= menor) {
      answer += colors[i];
      val = colors[i];
      for (int p = i + 1, k = j + 1, t = corrimientos; t && k < n;
           ++i, ++p, ++k, ++j, i %= n) {
        if (colors[p] <= val)
          break;
        else {
          answer += val;
          --t;
        }
      }
    } else {
      answer += menor;
      ++corrimientos;
    }
  }
  cout << answer << endl;
  return 0;
}
