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 repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
typedef long long ll;
typedef long double lb;
typedef pair<int,int> P;
const int MOD = 1000000007;
const ll INF = 1e16;
const double EPS = 1e-10;
const double PI = 3.141592653589793;
template<class T> inline bool chmin(T& a, T b){if(a>b){a = b;return 1;}return 0;}
template<class T> inline bool chmax(T& a, T b){if(a<b){a = b;return 1;}return 0;}
const pair<int,int> fd[] = {make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)}; 
#define int long long
#define double long double

////////////////////////////////////////////////////////////////////
ll modpow(ll a,  ll n, const ll &MOD = MOD){
  ll ret = n == 0 ? 1 : modpow(a, n/2, MOD);
  (ret *= ret) %= MOD;
  if(n%2)((ret *= a) %= MOD);
  return ret;
}
///////////////////////___modinv___////////////////////
ll modinv(ll d, const ll &MOD = MOD){
  return modpow(d, MOD-2, MOD);
}
///////////////////////___prime_factor___////////////////////
map<int, int> prime_factor(int n){
  map<int, int> ret;
  for(int i = 2;i*i <= n;i++){
    while(n % i == 0){
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1){
    ret[n] = 1;
  }
  return ret;
}

////////////////////////////////////////////////////////////////////

int n,x,vec[2100];P a[2100];
signed main(){
  cin>>n>>x;
  rep(i,n)cin>>a[i].first, a[i].second = i;
  sort(a, a+n);
  int ans = INF*100LL;
  rep(i,n+1){
    int sum = 0;
    rep(j,n)vec[j] = 0;
    rep(j,n){
      int k = a[j].second;
      while(vec[k%n] == 0 && k <= a[j].second+i){
        vec[k%n] = a[j].first;
        k++;
      }
    }
    rep(j,n)sum += vec[j];
    sum += x*i;
    chmin(ans, sum);
  }
  cout << ans << endl;
}


