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
//2018.9.29 by ljz
#include<bits/stdc++.h>
using namespace std;
#define res register LL
#define LL long long
#define inf 0x3f3f3f3f
#define eps 1e-15
inline LL read(){
    res s=0;
    bool w=0;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=1;ch=getchar();}
    while(ch>='0'&&ch<='9')s=s*10+ch-'0',ch=getchar();
    return w?-s:s;
}
inline void _swap(res &x,res &y){
    x^=y^=x^=y;
}
inline LL _abs(const res &x){
    return x>0?x:-x;
}
inline LL _max(const res &x,const res &y){
    return x>y?x:y;
}
inline LL _min(const res &x,const res &y){
    return x<y?x:y;
}
const LL N=2e3+10;
namespace MAIN{
    LL n,x;
    LL a[N],b[N][N],ans=inf;
    inline void MAIN(){
        n=read(),x=read();
        for(res i=1;i<=n;i++)a[i]=read();
        for(res i=1;i<=n;i++){
            b[i][i]=a[i];
            for(res j=i+1;j<=n;j++)b[i][j]=_min(b[i][j-1],a[j]);
        }
        for(res i=0;i<=n;i++){
            res ret=i*x;
            for(res j=1;j<=n;j++){
                res k=j-i;
                if(k<=0)k+=n,ret+=_min(b[k][n],b[1][j]);
                else ret+=b[k][j];
            }
            ans=_min(ans,ret);
        }
        printf("%lld\n",ans);
    }
}
int main(){
    MAIN::MAIN();
    return 0;
}