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
#include <bits/stdc++.h>

using namespace std;

long long n, x, menor = LONG_MAX, answer = 0, val;
long long colors[2010];
bool ready[2010];

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    colors[0] = 0;
    cin >> n >> x;
    int inicial;

    for(int i = 0; i < n; ++i){
        cin >> colors[i];
        if(colors[i] < menor){
            menor = colors[i];
            inicial = i;
        }
    }

    long long real = 200000000000000;
    for(int l = 0; l < n; ++l){
        menor = color[l];
        answer = menor;
        int corrimientos = 0;
        menor += x;
        for(int i = (l + 1) % n, j = 1; j < n; ++i, ++j, i %= n){
            if(colors[i] <= menor){
                answer += colors[i];
                val = colors[i];
                for(int p = i + 1, k = j + 1, t = corrimientos; t && k < n; ++i, ++p, ++k, ++j, i %= n){
                    if(colors[p] <= val)
                        break;
                    else{
                        answer += val;
                        --t;
                    }
                }
            }
            else{
                answer += menor;
                ++corrimientos;
            }
        }
        menor -= x;
        real = min(answer, real);
    }


    cout << real << endl;


    return 0;
}