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
#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
//const ll mod=998244353;
const ll mod=1000000007;
ll GCD(ll a,ll b) {return b ? GCD(b,a%b):a;}
ll LCM(ll c,ll d){return c/GCD(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int main(){
    ll n,x;
    cin>>n>>x;
    V<ll> a(n);
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
  reverse(a.begin(),a.end());
    V<P> res(n,{inf,inf});
    ll ans=inf;
    for(ll i=0;i<n;i++){
        for(ll j=0;j<n;j++){
            int id=(i+j)%n;
            if(res[id].fi+res[id].se>a[i]+j*x){
                res[id].se=j*x;
                res[id].fi=a[i];
            }
            if(res[id].fi+res[id].se==a[i]+j*x && res[id] > j*x){
              res[id].se = j*x;
              res[id].fi = a[i];
            }
        }
    }
       ll sum=0,tmp=0;
        for(ll j=0;j<n;j++){
            chmax(tmp,res[j].se);
            sum+=res[j].fi;
        }
        cout<<sum+tmp<<"\n";
}
