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
    //   Sourav Verma IPG_2013  ABV_IIITMG Task @ Codeforces & Atcoder

#include<bits/stdc++.h>

#define ll                    long long int
#define loop(i,a,b)           for(int i=(int)a;i<(int)b;++i)
#define rloop(i,a,b)          for(int i=(int)a;i<=(int)b;++i)
#define loopr(i,a,b)          for(int i=(int)a;i>=(int)b;--i)
#define count_1(n)            __builtin_popcountll(n)
#define pb                    push_back
#define mset(a,b,c)           loop(i,0,b) a[i]=c
#define F                     first
#define S                     second
#define mp                    make_pair
#define clr(x)                x.clear()
#define MOD                   1000000007
#define MAX                   1e9
#define MIN                   -1e9
#define itoc(c)               ((char)(((int)'0')+c))
#define vi                    vector<int>
#define vvi                   vector<vi>
#define pll                   pair<ll,ll>
#define pii                   pair<int,int>
#define all(p)                p.begin(),p.end()
#define mid(s,e)              (s+(e-s)/2)
#define mini                  INT_MIN
#define sv()                  int t; scanf("%d",&t); while(t--)
using namespace std;
ll extgcd(ll a,ll b,ll& x,ll& y){if(b==0){x=1;y=0;return a;}else{int g=extgcd(b,a%b,y,x);y-=a/b*x;return g;}}
ll lcm(int x, int y) { return x * 1ll * y / (int)__gcd(x, y); }
ll hcf(ll a, ll b) {if(b>a) return (hcf(b, a)); if(a%b==0) return b; return (hcf(b, a%b));} 
ll modpow(ll a,ll b) {ll res=1;a%=MOD;for(;b;b>>=1){if(b&1)res=res*a%MOD;a=a*a%MOD;}return res;}


int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int n,x; cin>>n>>x;
    vector<vi >a(n,vi(n,pow(10,9)+5));
    loop(i,0,n)  cin >>a[i][0];
    loop(i,0,n){
        loop(j,1,n){
            int m=(i+j)%n;
            if(a[i][0]<a[m][j]) a[m][j]=a[i][0];
        }
    }
    loop(i,0,n) {
        int min=a[i][0];
        loop(j,1,n) {
            if(min<a[i][j]) a[i][j]=min;
            else min=a[i][j];
        }
    }
    ll sum=0;
    loop(i,0,n) sum+=a[i][0];
    loop(i,1,n){
        ll sum1=0;
        loop(j,0,n) sum1+=a[j][i];
        if((sum1+i*x)<sum)  sum=sum1+i*x;
    }
    printf("%lld\n",sum);
    return 0;
}