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<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define REP(i, limit) for(int i=0; i< limit; ++i)
#define LINT long long

int main(){
  int N;
  LINT x;
  cin>>N>>x;
  LINT a[N];
  REP(i, N)cin>>a[i];
  LINT cost[N];
  LINT mahou[N]={};
  REP(i, N) cost[i]=a[i];
  REP(i, N){
    REP(j, N){
      if(cost[j] > ((N+j-i)%N)*x + a[i]){
        cost[j] = x*((N+j-i)%N) + a[i];
        mahou[j] = (N+j-i)%N;
      }
    }
  }
  //REP(i, N)cout<<cost[i]<<endl;

  LINT ans=0;
  LINT mmax=0;
  REP(i, N){
    ans+=cost[i] - x*mahou[i];
    mmax = max(mmax, mahou[i]);
  }
  ans += mmax*x;
  cout<<ans<<endl;


  return 0;
}
