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 ld = long double;
using pint = pair<int, int>;
using pll = pair<ll, ll>;
using edge = pair<int, ll>;
using graph = vector<vector<int>>;
using WeightedGraph = vector<vector<edge>>;
const ll INF = 1LL << 60;
const int INF32 = 1 << 29;
const ll MOD = 1000000007;
int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  vector<pll> s(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    s[i] = {a[i], 0};
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      ll dist = i - j;
      if (dist < 0) {
        dist += n;
      }
      ll c1 = a[j] - s[i].first;
      ll c2 = (dist - s[i].second) * x;
      if (c1 + c2 < 0) {
        s[i].first = a[j];
        s[i].second = dist;
      } else if (c1 + c2 == 0) {
        if (s[i].second > dist) {
          s[i].first = a[j];
          s[i].second = dist;
        }
      }
    }
  }
  ll ans = 0, M = 0;
  for (int i = 0; i < n; ++i) {
    ans += s[i].first;
    M = max(M, s[i].second);
  }
  ans += M * x;
  cout << ans << endl;
}
