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;
import std.string;
import std.conv;
import std.typecons;
import std.algorithm;
import std.functional;
import std.bigint;
import std.numeric;
import std.array;
import std.math;
import std.range;
import std.container;
import std.ascii;
void times(alias pred)(int n) {
    foreach(i; 0..n) pred();
}
auto rep(alias pred, T = typeof(pred()))(int n) {
    T[] res = new T[n];
    foreach(ref e; res) e = pred();
    return res;
}

long INF = long.max/3;
void main() {
    int N;
    long x;
    readf("%d %d\n", &N, &x);
    long[] a = readln.split.to!(long[]);

    int[][] ary = new int[][](N, N);
    foreach(i; 0..N) {
        ary[i][] = -1;
        long s = 0;
        long m = INF;
        foreach(j; 0..N) {
            long t = a[(N+i-j)%N]+s;
            if (t < m) {
                m = t;
                ary[i][j] = j;
            }
            if (ary[i][j] < 0) ary[i][j] = ary[i][j-1];
            s += x;
        }
    }

    long ans = INF;
    foreach(t; 0..N) {
        long s = 0;
        foreach(i; 0..N) {
            s += a[(N+i-ary[i][t])%$];
        }
        ans = min(ans, s+t*x);
    }
    ans.writeln;
}
