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>
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
using namespace std;
int main() {
  int n, x;
  scanf("%d %d", &n, &x);
  vector<int> a;
  {
    int tmp;
    for (int i = 0; i < n; ++i) {
      scanf("%d", &tmp);
      a.push_back(tmp);
    }
  };
  int cost[2000];
  for (int i = 0; i < n; ++i) {
    cost[i] = a[i];
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (i > j) {
        cost[i] = min(cost[i], a[j] + x * (i - j));
      } else {
        cost[i] = min(cost[i], a[j] + x * ((n - j - 1) + i));
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    ans += cost[i];
  }
  printf("%lld\n", ans);
  return 0;
}
