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;
long long int n, x, arr[100000];
struct node {
  long long int top, cur, tut;
} bas[100000], ken[100000];
int main() {
  cin >> n >> x;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  for (long long int i = n + 1; i <= 2 * n; i++) arr[i] = arr[i - n];
  long long int maxi = -999999999999999, flag = 0;
  for (long long int j = 1; j <= n; j++) {
    long long int mini = 999999999999999;
    long long int cur = 0, tutgac = 0, tut = 0;
    for (long long int i = j + 1; i <= n + j - 1; i++) {
      tutgac = abs(n - i + j) % n;
      if (mini > tutgac * x + arr[i])
        mini = tutgac * x + arr[i], cur = arr[i], tut = tutgac;
    }
    bas[j].top = mini;
    bas[j].cur = cur;
    bas[j].tut = tut;
  }
  long long int ans = 999999999999999;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int sum = 0;
    for (long long int j = 1; j <= n; j++) {
      if (bas[j].tut > i) {
        sum += arr[j];
        continue;
      }
      if (bas[j].top > arr[j]) {
        sum += arr[j];
        continue;
      }
      sum += bas[j].cur;
    }
    sum += i * x;
    ans = min(ans, sum);
  }
  cout << ans;
}
