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 mod = 1e9 + 7, mod1 = 1e9 + 9;
const long long N = 2005;
long long a[N], mark[N], mark2[N];
long long MAX = 30000000000;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long n, x;
  cin >> n >> x;
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  }
  long long ans = 0;
  long long cnt = 0;
  while (cnt < n) {
    long long tmp = 0;
    for (long long i = 0; i < (n); i++)
      if (mark[i] == 0) {
        tmp += -a[(i + 1) % n] + a[i];
      }
    if (tmp > x) {
      ans += x;
      for (long long i = 0; i < (n); i++)
        if (mark[i] == 1) mark2[(i + 1) % n] = 1;
      swap(mark, mark2);
      for (long long i = 0; i < (n); i++) mark2[i] = 0;
    } else {
      long long ma = MAX, in;
      for (long long i = 0; i < (n); i++)
        if (mark[i] == 0 && a[i] < ma) {
          ma = a[i];
          in = i;
        }
      mark[in] = 1, ans += a[in], cnt++;
    }
  }
  cout << ans;
  return 0;
}
