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 pint = pair<int, int>;
using ll = long long;
using pll = pair<ll, ll>;
constexpr ll INF = 1e9 + 7;
constexpr ll MOD = 1e9 + 7;
template <typename T>
void vecin(vector<T> &v, int n = 0) {
  if (n == 0) n = v.size();
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
  }
}
template <typename T>
void vecout(vector<T> &v, int n = 0) {
  if (n == 0) n = v.size();
  for (int i = 0; i < n - 1; ++i) {
    cout << v[i] << " ";
  }
  cout << v[n - 1] << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ;
  int n, x;
  cin >> n >> x;
  vector<ll> a(n);
  vecin(a);
  ll ans = LLONG_MAX;
  vector<ll> b(n, LLONG_MAX);
  for (int i = (0), i_end_ = (n); i < i_end_; i++) {
    ll temp = i * x;
    for (int j = (0), j_end_ = (n); j < j_end_; j++) {
      b[j] = min(b[j], a[j - i >= 0 ? j - i : j - i + n]);
      temp += b[j];
    }
    ans = min(ans, temp);
  }
  cout << ans << endl;
}
