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];
  vector<vector<int>> B(N, vector<int>(N));
  for (int i = (int)(0); i < (int)(N); i++) {
    B[i][0] = i;
    long long mi = A[i];
    for (int j = (int)(1); j < (int)(N); j++) {
      if (mi >= A[(i - j + N) % N] + x * j) {
        B[i][j] = (i - j + N) % N;
        mi = A[(i - j + N) % N] + x * j;
      } else {
        B[i][j] = B[i][j - 1];
      }
    }
  }
  long long ret = 1LL << 55;
  for (int j = (int)(0); j < (int)(N); j++) {
    long long sum = 0;
    for (int i = (int)(0); i < (int)(N); i++) sum += A[B[i][j]];
    sum += x * j;
    ret = min(ret, sum);
  }
  output(ret, 0);
  return 0;
}
