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>
template <typename T>
void print_to(std::ostream &os, const char *, const char *tail, const T &fst) {
  os << fst << tail;
}
template <typename Fst, typename... Rst>
void print_to(std::ostream &os, const char *del, const char *tail,
              const Fst &fst, const Rst &...rst) {
  os << fst << del;
  print_to(os, del, tail, rst...);
}
template <typename Iter>
void print_to_(std::ostream &os, const char *del, const char *tail, Iter bgn,
               Iter end) {
  for (Iter it = bgn; it != end;) {
    os << *it;
    if (++it != end)
      os << del;
    else
      os << tail;
  }
}
template <typename Fst, typename... Rst>
void println(const Fst &fst, const Rst &...rst) {
  print_to(std::cout, "\n", "\n", fst, rst...);
}
template <typename Fst, typename... Rst>
void print(const Fst &fst, const Rst &...rst) {
  print_to(std::cout, " ", "\n", fst, rst...);
}
template <typename Iter>
void println_(Iter bgn, Iter end) {
  print_to_(std::cout, " ", "\n", bgn, end);
}
using namespace std;
const int inf = (int)1e9;
const int dy[] = {1, 0, -1};
const int dx[] = {1, 0, -1};
int n, m, h, w;
string s;
long long x, cost[2001], a[2001];
long long cont[2001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cost[i] = a[i];
  for (int i = 0; i < n; i++) cont[i] = inf;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (j > i) {
        cost[i] = min(cost[i], a[j] + x * (i + n - j));
        if (cost[i] == a[j] + x * (i + n - j))
          cont[i] = min(cont[i], (long long)i + n - j);
      } else {
        cost[i] = min(cost[i], a[j] + x * (i - j));
        if (cost[i] == a[j] + x * (i - j))
          cont[i] = min(cont[i], (long long)i - j);
      }
    }
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    if (cont[i] == inf)
      sum += cost[i];
    else
      sum += cost[i] - cont[i] * x;
  }
  long long maxcont = 0;
  for (int i = 0; i < n; i++) {
    if (cont[i] == inf) continue;
    maxcont = max(maxcont, cont[i]);
  }
  cout << sum + x * maxcont << endl;
  return 0;
}
