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;
int main() {
  int n;
  long long int x;
  cin >> n >> x;
  long long int a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long int b[n], c[n];
  for (long long i = 0; i < n; i++) b[i] = 10000000000000000LL;
  for (long long i = 0; i < n; i++) c[i] = 10000000000000000LL;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      if ((a[j] + x * ((i - j + n) % n)) < b[i] ||
          ((a[j] + x * ((i - j + n) % n)) == b[i] &&
           c[i] > ((i - j + n) % n))) {
        b[i] = (a[j] + x * ((i - j + n) % n));
        c[i] = ((i - j + n) % n);
      }
    }
  }
  long long int d[n];
  for (long long i = 0; i < n; i++) d[i] = b[i] - c[i];
  long long int m = 0;
  for (long long i = 0; i < n; i++) {
    if (m < d[i]) m = d[i];
  }
  for (long long i = 0; i < n; i++) m += c[i];
  cout << m << endl;
  return 0;
}
