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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    ArrayList<Integer>[] graph;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long x = sc.nextLong();
        long[] a = new long[n];
        for(int i=0;i<n;i++) a[i]=sc.nextLong();
        long[][] dp = new long[n][n];
        for(int i=0;i<n;i++) Arrays.fill(dp[i],Long.MAX_VALUE/10);
        for(int i=0;i<n;i++) dp[i][i]=a[i];
        for(int pass=0;pass<2;pass++){
            for(int i=0;i<n;i++) dp[i][i] = Math.min(dp[i][i],x+dp[(i-1+n)%n][(i-1+n)%n]);
        }
        for(int d=1;d<n;d++){
            for(int pass=0;pass<2;pass++){
                for(int start=0;start<n;start++){
                    long left = a[start]+dp[(start+1)%n][(start+d)%n];
                    long right = a[(start+d)%n]+dp[start][(start+d-1)%n];
                    long shift = x + dp[(start-1+n)%n][(start+d-1)%n];
                    long cur = Math.min(left,Math.min(right,shift));
                    dp[start][(start+d)%n]=Math.min(dp[start][(start+d)%n],cur);
                }
            }
        }
        long ans = Long.MAX_VALUE/10;
        for(int start=0;start<n;start++){
            ans = Math.min(ans,dp[start][(start+n-1)%n]);
        }
        System.out.println(ans);
    }
}
