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 <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;i++)
#define rep(i,n) for(ll i=0;i<n;i++)
#define vecll vector<ll>
using namespace std;
int in() {int x;scanf("%d",&x);return x;}
ll lin() {ll x;scanf("%lld",&x);return x;}
int main(){
    int n=in();
    ll x=lin();
    vecll a(n);
    rep(i,n)a[i]=in();
    vector<vector<ll>> b(n,vector<ll>(n));
    rep(i,n){
        b[i][0]=a[i];
        rep2(j,1,n-1) b[i][j]=min(b[i][j-1],a[(i-j+n)%n]);
    }
    ll ans=LONG_MAX;
    rep(k,n){
        ll tmp=x*k;
        rep(i,n) tmp+=b[i][k];
        ans=min(ans,tmp);
    }
    cout<<ans<<endl;
}