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;
long N,x,a[2000][2000],ans;//a[魔法を唱えられる回数i][スライムの色j]は魔法をi回以下唱えたときにスライムjを得るための最小時間
int main(){
  cin>>N>>x;
  for(int i=0;i<N;ans+=a[0][i++])cin>>a[0][i];//このfor文が終了するとansはaの合計値でこれは魔法を0回唱えたときに相当する
  for(int i=1;i<N;i++){//iは魔法を唱えられる回数
    long s=i*x;//魔法を唱えるのに必要な合計時間
    for(int j=0;j<N;s+=a[i][++j%N])a[i][(j+1)%N]=min(a[i-1][(j+1)%N],a[i-1][j]);//二次元の表を書くと意味がわかりやすい
    ans=min(ans,s);
  }
  cout<<ans<<endl;
}
