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;

int main(){
    long long n,x;
    cin>>n>>x;
    long long a[n];
    long long ans=0,m=1e10,p;
    long long po[n]={};
    for(int i=0;i<n;i++){
        cin>>a[i];
        if(a[i]==x) po[i]++;
        if(a[i]<m) m=a[i],p=i;
    }

    long long tmp=a[p];

    
    for(int i=p;i<n;i++){
        if(i!=n-1 && a[i+1]>=tmp+x) a[i+1]=x+tmp;
        else if(i!=n-1) tmp=a[i+1];
        if(i==n-1 && a[0]>tmp+x) a[0]=x+tmp;
        else if(i==n-1) tmp=a[0];
    }
    for(int i=0;i<p;i++){
        if(i!=n-1 && a[i+1]>=tmp+x) a[i+1]=x+tmp;
        else if(i!=n-1) tmp=a[i+1];
        if(i==n-1 && a[0]>tmp+x) a[0]=x+tmp;
        else if(i==n-1) tmp=a[0];
    }
    tmp=0;
    for(int i=p;i<n;i++) {
        ans+=a[i];
        if(a[i]!=x) tmp=a[i];
        if(po[i]!=0) tmp=a[i];
        else if(a[i]==x) ans+=tmp;
    }

    for(int i=0;i<p;i++){
        ans+=a[i];
        if(a[i]!=x) tmp=a[i];
        if(po[i]!=0) tmp=a[i];
        else if(a[i]==x) ans+=tmp;
    }
        
    cout<<ans;

}