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;
class SparseTable {
  vector<vector<int> > st;

 public:
  void init(vector<int> v) {
    int b = 0;
    for (; (1 << b) <= v.size(); b++)
      ;
    st.assign(b, vector<int>(1 << b));
    for (int i = 0; i < v.size(); i++) st[0][i] = v[i];
    for (int i = 1; i < b; i++) {
      for (int j = 0; j + (1 << i) <= (1 << b); j++) {
        st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
      }
    }
  }
  int query(int l, int r) {
    assert(l < r);
    int b = 32 - __builtin_clz(r - l) - 1;
    return min(st[b][l], st[b][r - (1 << b)]);
  }
  SparseTable() {}
  SparseTable(vector<int> v) { init(v); }
};
int N, X;
int main() {
  cin >> N >> X;
  vector<int> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  SparseTable st(a);
  long long ret = (1LL << 61);
  for (int i = 0; i < N; i++) {
    long long sum = 0;
    for (int j = 0; j < N; j++) {
      int l = j - i;
      if (l >= 0)
        sum += st.query(l, j + 1);
      else
        sum += min(st.query(0, j + 1), st.query(l + N, N));
    }
    sum += i * X;
    ret = min(ret, sum);
  }
  cout << ret << endl;
  return 0;
}
