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 <int POS, class TUPLE>
void deploy(std::ostream &os, const TUPLE &tuple) {}
template <int POS, class TUPLE, class H, class... Ts>
void deploy(std::ostream &os, const TUPLE &t) {
  os << (POS == 0 ? "" : ", ") << get<POS>(t);
  deploy<POS + 1, TUPLE, Ts...>(os, t);
}
template <class T, class U>
std::ostream &operator<<(std::ostream &os, std::pair<T, U> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::vector<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T>
std::ostream &operator<<(std::ostream &os, std::set<T> &v) {
  int remain = v.size();
  os << "{";
  for (auto e : v) os << e << (--remain == 0 ? "}" : ", ");
  return os;
}
template <class T, class K>
std::ostream &operator<<(std::ostream &os, std::map<T, K> &make_pair) {
  int remain = make_pair.size();
  os << "{";
  for (auto e : make_pair)
    os << "(" << e.first << " -> " << e.second << ")"
       << (--remain == 0 ? "}" : ", ");
  return os;
}
using ll = long long;
ll cost[2020];
int start[2020];
ll C[2020];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int N;
  cin >> N;
  ll x;
  cin >> x;
  for (int i = 0; i < (N); i++) {
    cin >> cost[i];
    start[i] = i;
  }
  for (int i = 0; i < (N); i++) {
    for (int j = 0; j < (N); j++) {
      if (cost[i] > cost[j] + ((N + i - j) % N) * x) {
        cost[i] = cost[j] + ((N + i - j) % N) * x;
        start[i] = j;
      }
    }
  }
  memset(C, 0, sizeof(C));
  for (int i = 0; i < (N); i++) {
    C[start[i]] += 1;
  }
  ll maxc = 0;
  ll ans = 0;
  for (int i = 0; i < (N); i++) {
    if (C[i] > 0) {
      ans += C[i] * cost[i];
      maxc = max(maxc, C[i]);
    }
  }
  ans += (maxc - 1) * x;
  cout << ans << "\n";
  return 0;
}
