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 <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
// #include <boost/multiprecision/cpp_int.hpp> 

#define gc getchar//_unlocked
#define pc putchar//_unlocked
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define pp pair<int,int>
#define ppl pair<ll,ll>
#define bigint boost::multiprecision::cpp_int
#define finp ios_base::sync_with_stdio(0);cin.tie(0);
#define bc __builtin_popcountll
#define afor(i,a,b) for(int i=a;i<=b;++i)
#define bfor(i,a,b) for(int i=a;i>=b;--i)
#define vi vector<int>
#define vpp vector<pp>
#define vll vector<ll>

using namespace std;
using namespace __gnu_pbds;

char putnb[20];
void putn(ll n) {if(!n)pc('0');if(n<0)pc('-'),n=0-n;int pi=0;while(n)putnb[pi++]=(n%10)+'0',n/=10;while(pi)pc(putnb[--pi]);}
void sci(int *x) {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x = (int)((((*x)<<1) + ((*x)<<3)) + c - 48);}
void scll(ll *x)  {register char c = gc();*x = 0;for(; (c<48)||(c>57);c = gc());for(; (c>47)&&(c<58);c = gc())*x =  (ll)((((*x)<<1) + ((*x)<<3)) + c - 48);}
ll fp(ll a,ll b,ll c) {if(b==0)return 1%c; if(b==1)return a%c; ll ret=fp(a,b/2,c); ret=(ret*ret)%c; if(b&1)ret=(ret*a)%c; return ret;}

const ll mod=1e9 +7;
const ll mod2=1999999973;
const ll inf=1e18;
const int infs=1e9 + 1000;
const int N=100000;
const long double PI = acos(-1);

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

ll n,x;
ll A[2005];

int main()
{
  finp;
  cin>>n>>x;
  ll ans = 0;
  ll minv = inf;
  ll st;
  afor(i,1,n)
  {
    cin>>A[i];
    if(A[i] < minv)
    {
      minv = A[i];
      st = i;
    }
  }

  ll curr = st;

  while(1)
  {
    ans+= min(A[st]+x,A[curr]);
    curr= curr%n +1;
    if(curr == st)break;
  }



  cout<<ans;
  
return 0;
}