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 <typename T>
using V = vector<T>;
template <typename T, typename U>
using P = pair<T, U>;
using ll = long long;
using PL = P<ll, ll>;
using PS = P<string, ll>;
using VI = V<int>;
using VVI = V<VI>;
using VL = V<ll>;
using VVL = V<VL>;
using VS = V<string>;
using VB = V<bool>;
using VVB = V<VB>;
using VPL = V<PL>;
using VPS = V<PS>;
const ll INF = 1LL << 50;
template <typename T>
T square(T a) {
  return a * a;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N, x;
  cin >> N >> x;
  VL a(N);
  for (ll i = (0); i <= (N - 1); i++) {
    cin >> a[i];
  }
  ll ans = INF;
  for (ll t = (0); t <= (N - 1); t++) {
    ll cost = 0;
    for (ll i = (0); i <= (N - 1); i++) {
      ll subcost = a[i];
      for (ll j = (1); j <= (t); j++) {
        subcost = min(subcost, a[(i + j) % N]);
      }
      cost += subcost;
    }
    ans = min(ans, cost + x * t);
  }
  cout << ans << endl;
  return 0;
}
