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 <iostream>
#include <algorithm>
#include <queue>
#include <cassert>
#include <set>
using namespace std; typedef long long ll; const int MINF=-1e9; typedef pair<int,int> P;

int main() {
  ll n,x; cin>>n>>x;
  ll a[n]; for(int i=0;i<n;i++) cin>>a[i];
  if (n==1) {
    cout<<a[0]<<endl; return 0;
  }
  ll an=1e18;
  ll b[n];for(int i=0;i<n;i++) b[i]=a[i];
  for(ll i=0;i<=n;i++) {
    ll sm=0;
    for(int j=0;j<n;j++) {
      sm+=b[j];
    }
    an=min(an,sm+x*i);
    //cout<<i<<" "<<an<<endl;
    if (i==n) break;

    ll temp0=min(b[0],b[n-1]);
    for(int j=n-1;j>=1;j--) {
      b[j]=min(b[j-1],b[j]);
    }
    b[0]=temp0;
    /for(int j=0;j<n;j++) cout<<b[j]<<" "; cout<<endl;
  }
  cout<<an<<endl;
}