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 argc, char *argv[]) {
  long N, x, as[(2000)];
  scanf("%ld %ld", &N, &x);
  for (int i = 0; i < N; i++) {
    scanf("%ld", &as[i]);
  }
  long time_catches[(2000)], time_magics[(2000)];
  for (int i = 0; i < N; i++) {
    time_catches[i] = (1000000000000000000L);
    time_magics[i] = (1000000000000000000L);
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      const long time_catch = as[j];
      const long time_magic = (i - j >= 0 ? i - j : i - j + N) * x;
      if (time_catch + time_magic < time_catches[i] + time_magics[i] ||
          (time_catch + time_magic == time_catches[i] + time_magics[i] &&
           time_magic > time_magics[i])) {
        time_catches[i] = time_catch;
        time_magics[i] = time_magic;
      }
    }
  }
  long ans_time_catch = 0, ans_time_magic = 0;
  for (int i = 0; i < N; i++) {
    ans_time_catch += time_catches[i];
    ans_time_magic = max(ans_time_magic, time_magics[i]);
  }
  printf("%ld\n", ans_time_catch + ans_time_magic);
  return 0;
}
