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() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  int sum = 0;
  int maxch = 0;
  for (int i = 0; i < n; ++i) {
    map<int, int> m1;
    map<int, int> m2;
    m1.insert(make_pair(a[i], i));
    m2.insert(make_pair(a[i], 0));
    for (int j = 1; j < n; ++j) {
      int loop;
      if (i + j >= n)
        loop = i + j - n;
      else
        loop = i + j;
      m1.insert(make_pair(a[loop] + x * (n - j), loop));
      m2.insert(make_pair(a[loop] + x * (n - j), n - j));
    }
    sum += a[m1.begin()->second];
    maxch = max(maxch, m2.begin()->second);
  }
  sum += maxch * x;
  cout << sum << endl;
  return 0;
}
