Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);
//cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll inf=1000000000000000000;
double pi=2*acos(0);
#define rep(i,m,n) for(int i=m;i<n;i++)
#define rrep(i,n,m) for(int i=n;i>=m;i--)
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
int main(){
    ll n,x;cin>>n>>x;
    ll a[n];rep(i,0,n)cin>>a[i];
    ll cnt[n];
    fill(cnt,cnt+n,0);
    rep(i,0,n){
        ll num;
        ll mi=inf;
        rep(j,i+1,n){
            if(a[j]+x*(i+n-j)<mi){
                mi=a[j]+x*(i+n-j);
                num=j;
            }
        }
        rep(j,0,i+1){
            if(a[j]+x*(i-j)<mi){
                mi=a[j]+x*(i-j);
                num=j;
            }
        }
        //cout<<mi<<" "<<num<<endl;
        cnt[num]++;
    }
    ll ma=0;
    ll ans=0;
    rep(i,0,n){
        ans+=a[i]*cnt[i];
        ma=max(ma,cnt[i]);
    }
    cout<<ans+x*(ma-1)<<endl;
}