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 a[2048];
long long b[2048];
long long c[2048];
int main() {
  long long n, x, k, l = 0;
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long m = 10000000001;
  for (int i = 0; i < n; i++)
    if (a[i] < m) {
      m = a[i];
      k = i;
    }
  for (int i = n - 1; i >= 0; i--) {
    long long y = a[(k + i) % n];
    long long p = 0;
    m = y * (n - i) + x * (n - i - 1);
    long long d = n;
    long long z = 0;
    for (int j = 0; j < l; j++) {
      z += (d - b[j]) * c[j] + (d - b[j] - 1) * x;
      d = b[j];
      long long w = (d - i) * y + (d - i - 1) * x;
      if (z + w < m) {
        m = z + w;
        p = j + 1;
      }
    }
    b[p] = i;
    c[p] = y;
    l = p + 1;
  }
  long long d = n;
  long long z = 0;
  for (int i = 0; i < l; i++) {
    z += (d - b[i]) * c[i] + (d - b[i] - 1) * x;
    d = b[i];
  }
  cout << z << endl;
}
