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[2 * N];
		int[] a2 = new int[N];
		for (int i = 0; i < N; i++) {
			a[i] = s.nextInt();
			a[N + i] = a[i];
		}

		int min = a[0];
		int minno = 0;
		for (int i = 1; i < N; i++) {
			min = Math.min(min, a[i]);
			if (min == a[i]) {
				minno = i;
			}
		}

		long t = a[minno];
		int count = 0;
		int countm = 0;
		a2[minno] = a[minno];

		for (int i = minno + 1; i < minno + N; i++) {
			if (a[i] <= t + x) {
				count = 0;
				t = a[i];
			} else {
				count++;
				t = t + x;
			}
			if (count > countm) {
				countm = count;
			}
		}

		for (int i = 0; i < N; i++) {
			min = a[i];
			if (i >= countm) {
				for (int j = i - countm; j < i; j++) {
					min = Math.min(min, a[j]);
				}
				a2[i]=min;
			}else{
				for(int j=N+i-countm;j<i+N;j++){
					min = Math.min(min, a[j]);
				}
				a2[i]=min;
			}
		}

		long result = (long) countm * x;
		for (int i = 0; i < N; i++) {
			result += (long) a2[i];
		}
		System.out.println(result);
	}
}