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<iomanip>
#include<cstdio>
#include<string>
#include<cstring>
#include<climits>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<deque>
#include<tuple>
#include<list>
#include<unordered_map>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vii vector<int>
#define vll vector<ll>
#define lb lower_bound
#define pb push_back
#define mp make_pair
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define INF (1 << 30) - 1
#define LLINF (1LL << 61) - 1
// #define int ll
using namespace std;
const int MOD = 1000000007;
const int MAX = 510000;
 
 
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,x,a[3999],b[3999][3999],ans=INF;
    cin>>n>>x;
    rep(i,n) cin>>a[i];
    rep(i,n) b[i][0]=a[i];
    rep(i,n){
        ll sec=0;
        
        rep(j,n){
         if(i==0){
                sec+=a[j];
                continue;
          }
         ll s=j-i;
         if(j-i<0) s+=n;
         b[j][i]=min(b[j][i-1],a[s]);, 
         sec+=min(b[j][i-1],a[s]);
        }
        ans=min(ans,sec+i*x);
}
    cout<<ans<<endl;
    return 0;
}