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, a[2020], mi[2020][3], t, ans, x;
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) mi[i][1] = a[i];
  for (int i = 2; i <= 2 * n * n + 2; i++) {
    if (t >= mi[((i - 2) % n + 1)][2] + 1 &&
        mi[((i - 2) % n + 1)][1] < mi[((i - 1) % n + 1)][1]) {
      mi[((i - 1) % n + 1)][1] = mi[((i - 2) % n + 1)][1];
      mi[((i - 1) % n + 1)][2] = mi[((i - 2) % n + 1)][2] + 1;
    }
    if (mi[((i - 2) % n + 1)][1] + x < mi[((i - 1) % n + 1)][1]) {
      t = mi[((i - 2) % n + 1)][2] + 1;
      mi[((i - 1) % n + 1)][1] = mi[((i - 2) % n + 1)][1];
      mi[((i - 1) % n + 1)][2] = mi[((i - 2) % n + 1)][2] + 1;
    }
  }
  for (int i = 1; i <= n; i++) ans += mi[i][1];
  cout << ans + x * t;
}
