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 Int = long long;
typedef pair<int,int> P; typedef priority_queue<int> Pr; typedef vector<int> Vi; typedef vector<P> Vp; typedef map<int,int> MP; typedef vector<Int> VI;
const int mod = 1000000007; const Int INF = 1145141919810893931;
#define END {cout<<ans<<'\n'; return 0;}
#define ALL(v) v.begin(),v.end()
template<class T> inline bool cmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool cmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main(){
  cin.tie(nullptr); ios::sync_with_stdio(false);

  Int n,k,mz=0,ans=0; cin>>n>>k; Int a[n],c[n]={}; for (int i = 0; i < n; i++)cin>>a[i];
  VI d;
  for (int i = 0; i < n; i++){
    d.assign(n,0);
    for (int j = 0; j < n; j++){
      if(!j)d[i]=a[i];
      else d[(i-j+n)%n]=a[(i-j+n)%n]+k*j;
    }
    //for(auto k:d)cout<<k<<' '; cout<<endl;
    int use=distance(d.begin(),min_element(ALL(d)));
    c[use]++;
    cmax(mz,(use-i+n)%n);
  }
  for (int i = 0; i < n; i++)ans+=a[i]*c[i];
  cout<<ans+k*mz<<'\n';
}