Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
 
#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)
 
#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=(int)(a)-1;i>=(int)(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)
 
#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
 
// #define DEBUG
 
#ifdef DEBUG
    #define dump(...) fprintf(stderr, __VA_ARGS__)
#else
    #define dump(...)
#endif
 
template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}
 
using namespace std;
using ll=long long;
using vi=vector<int>;
using vll=vector<ll>;
 
const double EPS = 1e-10;
const double PI = acos(-1.0);
const ll inf =1LL << 62;
const ll mod=1000000007LL;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
 
 
ll extgcd(ll a,ll b,ll& x,ll& y){x=1,y=0;ll g=a;if(b!=0) g=extgcd(b,a%b,y,x),y-=a/b*x;return g;}
ll ADD(const ll &a, const ll &b,const ll &mod) { return (a+b)%mod;}
ll SUB(const ll &a, const ll &b,const ll &mod) { return (a-b+mod)%mod;}
ll MUL(const ll &a, const ll &b,const ll &mod) { return (1LL*a*b)%mod;}
ll DIV(const ll &a, const ll &b,const ll &mod) {ll x,y; extgcd(b,mod,x,y);return MUL(a,(x+mod)%mod,mod);}
 
random_device rd;
mt19937 mt(rd());
uniform_int_distribution<int> dice(1,6);
uniform_real_distribution<double> score(0.0,10.0);

int main(void){
    int n; cin >> n;
    ll x; cin >> x;
    vll a(n); for(auto & e : a) cin >> e;

    int mi = min_element(_all(a)) - begin(a);

    ll res = 0LL;
    ll cmin = a[mi];
    int cnt = 0;
    int mx_num = 0;
    rep(i, 1, n + 1){
        int ci = (mi + i) % n;
        int pi = (ci - 1 + n) % n;

        if(i == n){
            res += (cnt + 1) * cmin;
            chmax(mx_num, cnt);
            break;
        }

        if(chmin(a[ci], a[pi] + x)){
            cnt++;
        }
        else {
            res += (cnt + 1) * cmin;
            chmax(mx_num, cnt);
            cnt = 0;
            cmin = a[ci];
        }
    }
    res += mx_num * x;

    cout << res << endl;

    return 0;
}