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>
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);
  for (int i = (int)(0); i < (int)((N)); i++) {
    cin >> A[i];
    memo[i] = make_pair(A[i], 0);
  }
}
void solve() {
  for (int i = (int)(0); i < (int)((N)); i++)
    for (int j = (int)(0); j < (int)((N)); j++)
      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;
  for (int i = (int)(0); i < (int)((N)); i++) {
    ans += memo[i].first;
    chmax(max_diff, memo[i].second);
  }
  ans += max_diff * X;
  cout << ans << "\n";
}
}  // namespace solver
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  solver::init();
  solver::solve();
  return 0;
}
