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 N;
long long a[2001];
long long x;
long long now[2001];
vector<pair<long long, int> > v;
int main() {
  cin >> N >> x;
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
    v.push_back(pair<long long, int>(a[i], i));
  }
  sort(v.begin(), v.end());
  long long ans = 1e17;
  for (int i = 0; i < (N + 1); i++) {
    long long c = 0;
    for (int j = 0; j < (N); j++) {
      now[j] = -1;
    }
    for (int j = 0; j < (v.size()); j++) {
      int me = v[j].second;
      if (now[me] != -1) continue;
      c += v[j].first;
      now[me] = v[j].first;
      for (int k = 0; k < (i); k++) {
        if (now[(me + k + 1) % N] != -1) break;
        c += v[j].first;
        now[(me + k + 1) % N] = v[j].first;
      }
    }
    ans = min(c + i * x, ans);
  }
  cout << ans << endl;
  return (0);
}
