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>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
const long long INF = (1LL << 59);
using namespace std;

int main(){
    int n, x;
    cin >> n >> x;

    long long dp[2000][2] = {0};
    long long a[2000];
    rep(i,n) cin >> a[i];
    rep(i,n) dp[i][0] = INF;

    rep(i,n){
        int parent;
        rep(j,n){
            int diff;
            if(i <= j) diff = j - i;
            else if(i > j) diff = n - (i - j);

            if(dp[i][0] > diff * x + a[j]){
            //show(diff*x+a[j])
                dp[i][0] = diff * x + a[j];
                parent = j;
               //show(parent)
            }
        }
        //cout << i << ' ' << parent << endl;
        assert(parent < 2000)
        assert(parent >= 0)
        dp[parent][1]++;
    }

    long long sum = 0;
    rep(i,n) sum+=dp[i][1] * a[i];

    long long maxi = 0;
    rep(i,n) maxi = max(maxi, dp[i][1]);
    //rep(i,n) show(dp[i][1])

    if(maxi) sum+=(maxi - 1) * x;

    cout << sum << endl;
}
