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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n-1);i>=0;i--)
#define FOR(i,n,m) for(int i=n;i<=(int)(m);i++)
#define RFOR(i,n,m) for(int i=(int)(n);i>=m;i--)
#define all(x) (x).begin(),(x).end()
#define sz(x) int(x.size())
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
using namespace std;
vector<int> dy={1,0,-1,0};
vector<int> dx={0,1,0,-1};
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
int main(){
    int n;
    ll x;
    cin>>n>>x;
    vector<ll> a(n);
    rep(i,n) cin>>a[i];
    
    vector<ll> as(n);
    ll max_x=0;
    rep(i,n){
        ll now=LINF;
        int now_j=0;
        FOR(j,0,i){
            if(now>=(ll)a[j]+(i-j)*x){
                now=(ll)a[j]+(i-j)*x;
                now_j=j;
            }
        }
        FOR(j,i+1,n-1){
            if(now>=(ll)a[j]+(i+1)*x+(n-j-1)*x){
                now=(ll)a[j]+(i+1)*x+(n-j-1)*x;
                now_j=j;
            }
        }
        as[i]=a[now_j];
        max_x=max(max_x,now-as[i]);
    }
    
    ll ans=max_x;
    rep(i,n) ans+=as[i];
    cout<<ans<<endl;
}
