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)

/*
memo for solution
*/
#include <cstdlib> 
#include <numeric>
#include <cstring>
#include <cstdio> 
#include <cfloat> 
#include <map> 
#include <cassert>
#include <cmath> 
#include <climits> 
#include <utility> 
#include <set> 
#include <iostream> 
#include <memory> 
#include <string> 
#include <vector> 
#include <algorithm> 
#include <functional> 
#include <sstream> 
#include <complex> 
#include <stack> 
#include <queue> 
#include <iomanip>
#include<sys/time.h>
using namespace std; 
double e_time(void){static struct timeval now;gettimeofday(&now, NULL);return (double)(now.tv_sec  + now.tv_usec/1000000.0);}
#define REP(i,b,n) for(int i=b;i<n;i++) 
#define rep(i,n)      REP(i,0,n) 
#define pb push_back  
#define mp make_pair 
#define ALL(C)   (C).begin(),(C).end() 
#define fe(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr != (c).end();itr++)
#define BITSHIFT(X)     ( (1<<(X)) )
#define CONTAIN(S,X) ( ((S)&BITSHIFT(X)) != 0)
typedef complex<double>P; 
typedef long long ll; 
typedef unsigned long long ull; 
typedef pair<int,int> pii; 
typedef vector<int> vint;
template<class T> void vp(T &a,int p){rep(i,p)cout << a[i]<<" ";cout << endl;}  
template<class T> T ceilUp(const T& a,const T& b){return (a+b-1)/b;}
const ll mod =   1000000009;
inline ll modadd(ll a,ll b){a %= mod;b %= mod;return (a+b)%mod;}
inline ll modsub(ll a,ll b){a %= mod;b %= mod;a-=b;a%=mod;if (a<0)a+=mod;return a;}
inline ll modmul(ll a,ll b){a %= mod;b %= mod;return (a*b)%mod;}
inline ll modeq(ll a,ll b,ll m){a = (a%m+m)%m; b = (b%m+m)%m;return a == b;}
//ll modpow(ll n,ll p){if (p == 0)return 1;if (p == 1)return n;ll ret=mypow(n,p/2);ret=ret*ret%mod;if (p&1)ret=ret*n%mod;return ret;}
//ll getinv(ll a){return mypow(a,mod-2);}
ll mypow(ll n,ll p){
  if (p == 0)return 1;
  if (p == 1)return n;
  ll ret=mypow(n,p/2);
  ret=ret*ret;
  if (p&1)ret=ret*n;
  return ret;
}

const int N = 2000;
int table[N][N];

ll solve(int n, vector<ll>&in, int x){
  rep(i,n){
    rep(j,n){
      table[i][j] = in[(i-j+n)%n];
    }
    REP(j,1,n){
      table[i][j] = min(table[i][j], table[i][j-1]);
    }
  }

  ll ret = accumulate(begin(in), end(in), 0LL);
  rep(j,n){
    ll tmp = j * x;
    rep(i,n){
      tmp += table[i][j];
    }
    ret = min(ret, tmp);
  }
  return ret;
}
 
int main(){
  int n, x;
  cin >> n >> x;
  vector<ll> in(n);
  rep(i,n)cin>>in[i];
  cout << solve(n, in, x) << endl;
  return 0;
}