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 a[200005];
long long shifts[200005], mark[200005];
int main() {
  long long n, m, x, i, j;
  long long mn = 2000000200;
  scanf("%lld %lld", &n, &x);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  long long ans = 1000000002000000000ll;
  for (i = 0; i < n; i++) {
    long long temp = a[i];
    for (j = 0; j < n; j++) {
      shifts[j] = 0;
      mark[j] = 0;
    }
    mark[i] = 1;
    long long tot = 0, prev1 = i, l1 = 0;
    for (j = (i + 1) % n; j != i; j = (j + 1) % n) {
      if (a[j] > (x + a[i])) {
        temp += (a[prev1]);
        tot++;
      } else {
        prev1 = j;
        temp += a[j];
        mark[j] = 1;
        shifts[j] = tot;
        tot = 0;
      }
    }
    long long mn = tot;
    for (j = 0; j < n; j++) {
      if (mark[j] == 1) {
        mn = max(mn, shifts[j]);
      }
    }
    ans = min(ans, temp + mn * x);
  }
  printf("%lld\n", ans);
  return 0;
}
