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;
const int INF = INT_MAX;
const long long INFL = 1LL << 40;
const double PI = atan(1) * 4;
const long long MOD = pow(10, 9) + 7;
int dy[] = {0, 0, 1, -1, 0};
int dx[] = {1, -1, 0, 0, 0};
template <typename T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (T& x : vec) is >> x;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  }
  return os;
}
int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  cin >> a;
  ll sum = 0;
  ll maxX = 0;
  for (int i = 0; i < n; i++) {
    pair<long long ll, , long long ll, > mn = {INFL, 0};
    for (int j = 0; j < n; j++) {
      if (i >= j) {
        if (mn.first > a[j] + (i - j) * x) {
          mn = {a[j] + (i - j) * x, i - j};
        }
      } else {
        if (mn.first > a[j] + (n + i - j) * x) {
          mn = {a[j] + (n + i - j) * x, n + i - j};
        }
      }
    }
    sum += mn.first - x * mn.second;
    maxX = max(maxX, mn.second);
  }
  cout << sum + maxX * x << endl;
  return 0;
}
