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>
#define REP(i,n) for(int i = 0;i < (n);i++)
#define P pair<int,int>
#define pb push_back
#define mk make_pair
using namespace std;
const int INF = 1 << 20;

typedef long long ll;




int main(){
  ll n,x;
  cin >> n >> x;
  vector <ll> a;
  REP(i,n){
    ll b;
    cin >> b;
    a.pb(b);
  }
  bool d[n] = {0};
  ll now = 0;
  ll res = 0;
  while(1){
    bool f = true;
    REP(i,n){
      if(a[i] <= x && !d[(i+now)%n]){
	f = false;
	d[i+now] = true;
	res += a[i];
      }
    }
    now++;
    /*REP(i,n){
      cout << d[i] << "   " ;
      }*/
    //cout << endl;
    //cout << now << endl;
    res += x;
    if(f){
      res -= 2*x;
      break;
      
    }
  }
  cout << res << endl;
  return 0;
}
