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 a[3000], vis[3000];
int main() {
  int n, x;
  scanf("%d%d", &n, &x);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int mark = 1;
  for (int i = 1, j; i <= n; i++) {
    if (a[i] + x > a[1]) continue;
    for (j = i + 1; j <= n; j++)
      if (a[i] + x > a[j]) break;
    if (j == n + 1) {
      mark = i;
      break;
    }
  }
  int flag = mark, zz = mark;
  long long mi = 3000, sum = 0;
  long long ans = 0, z = 0;
  while (1) {
    if (mark == zz) z++;
    if (z == 2) break;
    if (vis[mark]) break;
    flag = mark + 1;
    if (flag == n + 1) flag = 1;
    long long res = 0;
    while (a[mark] + x <= a[flag]) {
      ans = ans + a[mark] + x;
      res++;
      vis[flag] = vis[mark] = 1, flag++;
      if (flag == n + 1) flag = 1;
    }
    if (res) mi = min(mi, res), sum++, ans += a[mark];
    mark = flag;
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) ans = ans + a[i];
  }
  if (sum) sum--;
  ans = ans - mi * sum;
  cout << ans << endl;
  return 0;
}
