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;
struct ST {
  long long a;
  int i;
};
bool operator<(const ST& a, const ST& b) {
  if (a.a == b.a) return a.i < b.i;
  return a.a < b.a;
}
int N;
long long solve(const vector<long long>& v, long long x) {
  long long ret = 0;
  for (int i = (0); i < (int)(v.size()); i++) {
    if (v[i] != 0) {
      long long a = 0;
      int j = (i + 1) % N;
      while (true) {
        if (v[j] == 0)
          a++;
        else
          break;
        j = (j + 1) % N;
      }
      ret += a * (v[i] + x) + v[i];
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  long long x;
  vector<ST> v;
  cin >> N >> x;
  long long ret = 0;
  for (int i = (0); i < (int)(N); i++) {
    long long a;
    cin >> a;
    v.push_back((ST){a, i});
    ret += a;
  }
  sort((v).begin(), (v).end());
  vector<long long> chk(N, 0);
  for (int i = (0); i < (int)(N); i++) {
    chk[v[i].i] = v[i].a;
    ret = min(ret, solve(chk, x));
  }
  cout << ret << endl;
  return 0;
}
