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>
using namespace std;

long b[2010][2010];
int main()
{
    cin.sync_with_stdio(0);
    cin.tie(0);

    long n, x; cin>>n>>x;
    long a[n+n];
    for(int i=0; i<n; ++i) { cin>>a[i]; a[n+i]=a[i]; }

    long s[n+n+1]={};
    for(int i=0; i<n+n; ++i) s[i+1]=s[i]+a[i];

    for(int i=0; i<n; ++i) {
        b[i][0]=0;
        for(int j=0; j<n; ++j)
            b[i][j+1]=b[i][j]+min(a[i], a[i+j]);
    }
    long res=1L<<60;
    for(int k=0; k<n; ++k) { // 魔法をk回使う
        for(int i=k; i<n+k; ++i) { // k回目の魔法でa[i]を作る
            long t=0;
            int p=i-k;
            //for(int j=0; j<=k; ++j) t+=min(a[p], a[p+j]);
            //for(int j=k+1; j<n; ++j) t+=a[p+j];
            t+=b[p][k+1];
            t+=s[p+n]-s[p+k+1];
            res=min(res, t+k*x);
        }
    }
    cout<<res<<endl;
}
