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("Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
//#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
//#pragma GCC target("avx2")  //Enable AVX
#include<bits/stdc++.h>
using namespace std;
#define int ll
#define all(a) begin(a),end(a)
#define F first
#define S second
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

const int mod=1e9+7;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int mul(int a,int b){
  return ((a)*1ll*(b))%mod;
}

void add(int &a,int b){
  a+=b;
  if(a>=mod)a-=mod;
}

int sub(int a,int b){
  a-=b;
  if(a<0){
    a+=mod;
  }
  return a;
}

int powz(int a,int b){
  int res=1;
  while(b){
    if(b&1){
      res=mul(res,a);
    }
    b/=2;
    a=mul(a,a);
  }
  return res;
}

template <typename A, typename B>
istream& operator>>(istream& input,pair<A,B>& x) {
  input>>x.F>>x.S;
  return input;
}

template <typename A>
istream& operator>>(istream& input,vector<A>& x) {
  for(auto& i:x)
    input>>i;
  return input;
}

template<typename A>
ostream& operator<<(ostream& output,vector<A>& x) {
  for(auto& i:x)
    output<<i<<' ';
  return output;
}

template<typename T>
vector<pair<T,int>> getvec(int n){
  vector<pair<T,int>>a(n);
  for(int i=0;i<a.size();i++){
    cin>>a[i].F;
    a[i].S=i;
  }
  return a;
}

void clr(auto &a,int n){
  a.clear();
  a.resize(n);
}

void unq(auto &a){
  sort(all(a));
  a.resize(unique(all(a))-a.begin());
}

const int N=200005;

int dp[2005][2005];

void solve(){
  int n,x;
  cin>>n>>x;
  vi a(n);
  cin>>a;
  if(n==1){
    cout<<a[0];
    return;
  }
  ll ans=1e18;
  for(int op=0;op<=n;op++){
    ll val=op*x;
    for(int i=0;i<n;i++){
      int mn=a[i];
      for(int j=0;j<op;j++){
	mn=min(mn,a[(i-j-1+n)%n]);
      }
      val+=mn;
    }
    ans=min(ans,val);
  }
  cout<<ans;
}

signed main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc=1;
  //cin>>tc;
  for(int _=0;_<tc;_++){
    // cout<<"Case #"<<_+1<<": ";
    solve();
    if(_!=tc-1){
      cout<<'\n';
    }
  }
}

