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 main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long int n, x;
  cin >> n >> x;
  vector<long long int> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  vector<long long int> cost(n);
  bool f = false;
  bool f2 = false;
  long long int old = 0;
  long long int count = 0;
  long long int maxcount = 0;
  long long int fcount = 0;
  for (int i = 0; i < (n); i++) {
    if (a[i] > x) {
      if (count == 0 || old == i - 1) count++;
      if (i != 0)
        cost[i] = cost[i - 1];
      else {
        f = true;
        f2 = true;
      }
      old = i;
    } else {
      if (count > 0) {
        if (f2) {
          fcount = count;
          f2 = false;
        }
        maxcount = max(maxcount, count);
        count = 0;
      }
      cost[i] = a[i];
    }
  }
  if (f) cost[0] = cost[n - 1];
  long long int ans = 0;
  if (maxcount > 0 || count > 0) {
    if (f)
      maxcount = max(fcount + count, maxcount);
    else if (count > 0)
      maxcount = max(maxcount, count);
    ans += maxcount * x;
  }
  for (int i = 0; i < (n); i++) ans += cost[i];
  cout << ans << endl;
  return 0;
}
