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 <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <limits.h>
#include <cmath>
#include <map>
using namespace std;
using P=pair<int,int>;
using ll=long long;
#define int long long
bool chmin(ll &a,ll b){
    if(a>b){
        a=b;
        return true;
    }else{
        return false;
    }
}
signed main(){
    ll n,x;
    cin>>n>>x;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    vector<int> cnt(n,0);
    int mx; 
    ll ret=0;
    for(int i=0;i<n;i++){
        ll plus=INT_MAX;
        int memo=-1;
        for(int j=0;j<n;j++){
            if(j<=i){
                if(chmin(plus,a[j]+x*(i-j))){
                    memo=j;
                }
            }else{
                if(chmin(plus,a[j]+x*(n-j+i))){
                    memo=j;
                }
            }
        }
        cnt[memo]++;
        if(memo<=i){
            mx=max(mx,i-memo);
        }else{
            mx=max(mx,(int)n-memo+i);
        }
    }
    for(int i=0;i<n;i++){
        ret+=a[i]*cnt[i];
        ret+=mx[i]*x;
    }
    cout<<ret<<endl;
}

