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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define len(v) ll(v.size())
#define fi first
#define se second

template <class T>
void cout_vec(const vector<T> &vec){
  for(auto itr:vec) cout<<itr<<' ';
  cout<<endl;
}

typedef pair<ll,ll> P;
const ll mod=1e9+7;
const ll inf=1e15;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n,x;
  cin>>n>>x;
  vector<ll> a(2*n);
  vector<vector<ll>> b(n,vector<ll>(n,inf));
  rep(i,n){
    cin>>a[i];
    a[i+n]=a[i];
  }
  FOR(i,n,2*n){
    b[i-n][0]=a[i];
    FOR(k,1,n){
      b[i-n][k]=min(b[i-n][k-1],a[i-k]);
    }
  }
  ll ans=inf;
  rep(k,n){//魔法を唱える回数をk回に固定
    ll cnt=k*x;
    rep(i,n){
      cnt+=b[i][k];
    }
    ans=min(ans,cnt);
  }
  cout<<ans<<endl;
}