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;
using ll = long long;
using Graph = vector<vector<int>>;
using WGraph = vector<vector<pair<int, ll>>>;
template <class T>
inline bool chmax(T &a, const T &b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
constexpr int dx[4] = {-1, 0, 0, 1}, dy[4] = {0, -1, 1, 0};
constexpr int INF = 1e9;
constexpr int MOD = 1e9 + 7;
constexpr long long LINF = 1e18;
constexpr double EPS = 1e-10;
constexpr double PI = M_PI;
void solve() {
  int n;
  ll x;
  cin >> n >> x;
  vector<ll> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a.at(i);
  }
  ll res = 0;
  int shift = 0;
  for (int i = 0; i < n; ++i) {
    ll mini = LINF;
    int ind = 0;
    for (int j = 0; j < n; ++j) {
      int s = (n + i - j) % n;
      if (chmin(mini, a.at(j) + x * s)) {
        ind = j;
      }
    }
    res += a.at(ind);
    chmax(shift, (n + i - ind) % n);
  }
  cout << res + x * shift << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  solve();
  return 0;
}
