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>

using namespace std;

#define debug(n) cerr << #n << ':' << n << endl;
#define dline    cerr << __LINE__ << endl;
#define rep(i,s,e) for(ll i = (ll)(s); i < (ll)(e); ++i)
#define per(i,s,e) for(ll i = (ll)(s)-1; i >= (ll)(e); --i)

using ll = long long;
template<class T, class U> using P = pair<T,U>;
template<class T> using Heap = priority_queue<T>;
template<class T> using heaP = priority_queue<T,vector<T>,greater<T>>;

template<class T>
bool ChangeMax(T&a,const T&b){
  if(a >= b) return false;
  a = b;    return true;
}

template<class T>
bool ChangeMin(T&a,const T&b){
  if(a <= b) return false;
  a = b;    return true;  
}

template<class T, size_t N, class U>
void Fill(T (&a)[N], const U&v){
    fill((U*)a,(U*)(a+N),v);
}

template<class T>
istream& operator >> (istream&is, vector<T>&v){
  for(auto&e:v)is >> e;
  return is;
}

int main(){
  int n,x; cin >> n >> x;
  vector<int> a(n); cin >> a;
  vector<int> v(n,1001001001);
  ll ans = 2010 * (1e+9);
  rep(i,0,n){
    ll now = i*x;   
    rep(i,0,n){
      ChangeMin(v[i], a[(i-k+n)%n]);
      now += v[i];
    }
    ChangeMin(ans, now);
  }
  cout << ans << endl;
  return 0;
}
