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 std.stdio, std.algorithm, std.conv, std.array, std.string, std.math, std.typecons, std.numeric;

void main()
{
    auto nx = readln.split.to!(long[]);
    auto N = nx[0];
    auto x = nx[1];
    auto as = readln.split.to!(long[]);

    long s, max_s;
    foreach (i; 0..N) {
        long t = long.max;
        foreach (j; 0..N) {
            if (i == j) {
                t = min(t, as[i]);
            } else if (j < i) {
                t = min(t, as[j] + max(i-j - max_s, 0) * x);
            } else {
                t = min(t, as[j] + max(i+N-j - max_s, 0) * x);
            }
        }
        s += t;
        long max_t;
        foreach (j; 0..N) {
            if (j < i && as[j] + max(i-j - max_s, 0) * x == t) {
                max_t = max(max_t, i-j);
            } else if (j > i && as[j] + max(i+N-j - max_s, 0) * x == t) {
                max_t = max(max_t, i+N-j);
            }
        }
        max_s = max(max_s, max_t);
    }
    writeln(s);
}