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.util.stream.Collectors;

public class Main {

    public static void main(String[] args) {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    // int N = scanner.nextInt();
    // scanner.nextLine();
    // int[] n = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    static void solve(Scanner scanner) {
        int N = scanner.nextInt();
        long X = scanner.nextLong();
        scanner.nextLine();
        int[] a = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();

        long[][] min = new long[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(min[i], Long.MAX_VALUE / 2);
            min[i][i] = a[i];
        }
        for (int i = 1; i < N; i++) {
            for (int l = 0; l < N; l++) {
                min[l][Math.floorMod(l + i, N)] = Math.min(min[l][Math.floorMod(l + i - 1, N)], a[Math.floorMod(l + i, N)]);
            }
        }

        long ans = Long.MAX_VALUE;
        for (int i = 0; i < N; i++) {
            long tmp = i * X;
            for (int j = 0; j < N; j++) {
                tmp += min[Math.floorMod(j - i, N)][j];
            }
            ans = Math.min(ans, tmp);
        }
        System.out.println(ans);
    }

}