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;
constexpr int MOD = 1000000007;
constexpr int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, -1, 1};
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "[";
  for (const auto &v : vec) {
    os << v << ",";
  }
  os << "]";
  return os;
}
void solve() {
  int N, x;
  cin >> N >> x;
  vector<long long> A(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  vector<long long> dist = A;
  int mi = min_element((A).begin(), (A).end()) - A.begin();
  long long cnt = 0, d = 0;
  for (int k = 1; k < N; k++) {
    if (dist[(mi + k) % N] > dist[(mi + k - 1) % N] + x) {
      dist[(mi + k) % N] = dist[(mi + k - 1) % N] + x;
      cnt++;
    } else {
      d += (1 + cnt) * cnt / 2 - cnt;
      cnt = 0;
    }
  }
  d += (1 + cnt) * cnt / 2 - cnt;
  long long sum = 0;
  for (int i = 0; i < N; i++) {
    sum += dist[i];
  }
  cout << sum - d * x << endl;
}
int main() {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(16);
  solve();
  return 0;
}
