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>

#ifdef _PRINTDEBUG
  #include "lib/printdebug.hpp"
#else
  #define printdebug(...) 1
#endif
#define I1000000007 (1000000007)
#define I998244353   (998244353)
#define IINF             (1<<28)
#define LINF           (1LL<<60)
#define ALL(a)         (a).begin(), (a).end()
#define rep(i, n)      REP(i, 0, (n))
#define repc(i, n)     REPC(i, 0, (n))
#define REP(i, n, m)   for (int i = (int)(n); i <  (int)(m); i++)
#define REPC(i, n, m)  for (int i = (int)(n); i <= (int)(m); i++)
#define REPCM(i, n, m) for (int i = (int)(n); i >= (int)(m); i--)
#define yesno(cond)       YESNO((cond), "Yes", "No")
#define YESNO(cond, a, b) cout << ((cond) ? (a) : (b)) << "\n"
template<class T> inline bool chmin(T& a, const T& b) {
  if (a > b) { a = b; return true; } else return false;
}
template<class T> inline bool chmax(T& a, const T& b) {
  if (a < b) { a = b; return true; } else return false;
}

namespace solver {
  using namespace std;
  using ll = long long;

  ll N, X;
  vector<ll> A;
  vector<pair<ll, ll> > memo;

  void init() {
    cin >> N >> X;
    A.resize(N);
    memo.resize(N);
    rep(i, N) {
      cin >> A[i];
      memo[i] = make_pair(A[i], 0);
    }
  }

  void solve() {
    rep(i, N) rep(j, N) if(i != j) {
      ll diff = j - i;
      if(diff < 0) diff += N;
      if(A[i]+diff*X < memo[j].first+memo[j].second*X) {
        memo[j].first = A[i];
        memo[j].second = diff;
      }
    }
    ll ans = 0;
    ll max_diff = 0;
    rep(i, N) {
      ans += memo[i].first;
      chmax(max_diff, memo[i].second);
    }
    ans += max_diff*X;
    cout << ans << "\n";
  }
}

int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  solver::init();
  solver::solve();
  return 0;
}