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 dx[8] = {1, 0, 0, -1, 1, 1, -1, -1};
int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};
int main() {
  int n;
  scanf("%d", &(n));
  ;
  long long x;
  SCANF_LL(x);
  long long aa[n * 3];
  for (int i = (0), loop_end_i = (n); i < (loop_end_i); i++) {
    long long p;
    SCANF_LL(p);
    aa[i] = aa[i + n] = aa[i + 2 * n] = p;
  }
  long long *a = aa + n;
  long long minans = LLONG_MAX;
  long long estim[n];
  int frm[n];
  fill(estim, estim + n, LLONG_MAX);
  for (int i = (0), loop_end_i = (n); i < (loop_end_i); i++) {
    for (int j = (0), loop_end_j = (n); j < (loop_end_j); j++) {
      long long nw = a[j - i] + x * i;
      if (nw < estim[j]) {
        frm[j] = j - i;
        estim[j] = nw;
      }
    }
    long long sm = i * x;
    for (int j = (0), loop_end_j = (n); j < (loop_end_j); j++) sm += a[frm[j]];
    ((minans) = min((minans), (sm)));
  }
  printf("%lld\n", minans);
}
