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>
const int MAXN = 2016;
int array[MAXN]{};
long long Ans, K;
int N, R;
int main() {
  scanf(" %d%lld", &N, &K);
  for (int i = 1; i <= N; ++i) scanf("%d", &array[i]);
  R = 0;
  for (int i = 1; i <= N; ++i) {
    long long cost = array[i];
    int ret = 0;
    for (int j = 1; j <= N; ++j)
      if (i != j) {
        if (j < i) {
          long long next = (array[j] + K * (i - j));
          if ((next < cost) || ((next == cost) && (i - j <= R))) {
            cost = array[j] + K * (i - j);
            R = std::max(i - j, R);
            ret = j;
          }
        }
        if (j > i) {
          long long next = (array[j] + K * (N - j + i));
          if ((next < cost) || ((next == cost) && (N - j + i <= R))) {
            cost = array[j] + K * (N - j + i);
            R = std::max(N - j + i, R);
            ret = j;
          }
        }
      }
    if (cost == array[i])
      Ans += cost;
    else
      Ans += array[ret];
  }
  Ans += 1LL * K * R;
  printf("%lld\n", Ans);
  return 0;
}
