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
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {

		Scanner s= new Scanner(System.in);
		int N = s.nextInt();
		int x=s.nextInt();
		int[] a = new int[N];
		for(int i=0;i<N;i++){
			a[i]=s.nextInt();
		}
		int[] a2=new int[2*N];
		for(int i=0;i<N;i++){
			a2[i]=a[i];
		}
		for(int i=N;i<2*N;i++){
			a2[i]=a[i-N];
		}
		
		int minno=0;
		int amin=a[0];
		for(int i=1;i<N;i++){
			if(a[i]<amin){
				amin=a[i];
				minno=i;
			}
		}		
		
		int cmax=0;
		int ccount=0;
		int time=a2[minno];
		for(int i=minno;i<minno+N;i++){
			if(a2[i]<time+x){
				cmax=Math.max(cmax, ccount);
				ccount=0;
				time=a2[i];
			}else{
				time+=x;
				ccount++;
			}
		}
		cmax=Math.max(cmax, ccount);
		
		long sum=0;
		for(int i=N;i<2*N;i++){
			 int min=100000000;
			 for(int j=i;j>=i-cmax;j--){
				 min=Math.min(a2[j],min );
			 }
			 sum+=min;
		}
	
		System.out.println(sum+x*cmax);
	}
}