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;
template <typename T>
inline void output(T a, int p) {
  if (p)
    cout << fixed << setprecision(p) << a << "\n";
  else
    cout << a << "\n";
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int N;
  long long x;
  cin >> N >> x;
  vector<long long> A(N);
  for (int i = (int)(0); i < (int)(N); i++) cin >> A[i];
  long long ret = 0;
  int xx = 0;
  for (int i = (int)(0); i < (int)(N); i++) {
    long long tmp = A[i];
    long long a = A[i];
    int tmp_x = 0;
    for (int j = i; j >= i - N; j--) {
      if (tmp > A[(j + N) % N] + x * (i - j)) {
        tmp = A[(j + N) % N] + x * (i - j);
        a = A[(j + N) % N];
        tmp_x = i - j;
      }
    }
    xx = max(xx, tmp_x);
    ret += a;
  }
  output(ret + xx * x, 0);
  return 0;
}
