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;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const long long mod = 1000000007;
const long long inf = long long(1e9);
const long long half_inf = long long(1e5);
const long long ll_inf = long long(1e9) * long long(1e9);
unordered_map<long long, long long> mp;
int main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (long long i = 0; i < (long long)(n); i++) cin >> a[i];
  long long m = 0;
  for (long long i = 0; i < (long long)(n); i++) {
    long long tmp = 0;
    long long tmp_sum = ll_inf;
    for (long long j = 0; j < (long long)(n); j++) {
      if (j == i)
        continue;
      else if (i > j) {
        if ((i - j) * x + a[j] < tmp_sum && (i - j) * x + a[j] < a[i]) {
          tmp_sum = (i - j) * x + a[j];
          tmp = i - j;
        }
      } else {
        if ((n - j - 1 + i + 1) * x + a[j] < tmp_sum &&
            (n - j - 1 + i + 1) * x + a[j] < a[i]) {
          tmp_sum = (n - j - 1 + i + 1) * x + a[j];
          tmp = (n - j - 1 + i + 1);
        }
      }
    }
    m = max(tmp, m);
  }
  long long ans = m * x;
  for (long long i = 0; i < (long long)(n); i++) {
    long long tmp = ll_inf;
    for (long long j = 0; j < (long long)(m + 1); j++) {
      if (i + j <= n - 1)
        tmp = min(tmp, a[i + j]);
      else
        tmp = min(tmp, a[j - n + i]);
    }
    ans += tmp == ll_inf ? 0 : tmp;
  }
  cout << ans << endl;
  return 0;
}
