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 n;
long long x;
long long a[2001];
int start[2001], skip[2001];
long long cost[2001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cost[i] = a[i];
    start[i] = i;
  }
  for (int i = 0; i < n; i++) {
    long long res = 0;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (i > j) {
        res = a[j] + x * (i - j);
      } else {
        res = a[j] + x * (n - j + i);
      }
      if (res < cost[i]) {
        cost[i] = res;
        start[i] = j;
        if (i > j)
          skip[i] = i - j;
        else
          skip[i] = n - j + i;
      }
    }
  }
  int maxskip = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    maxskip = max(maxskip, skip[i]);
    ans += a[start[i]];
  }
  cout << ans + x * maxskip << endl;
  return 0;
}
