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>
using namespace std;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
constexpr double EPS = 1e-9;
constexpr double PI = 3.1415926535897932;
constexpr int INF = 2147483647;
constexpr long long LINF = 1LL << 60;
constexpr long long MOD = 1000000007;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void Main() {
  long long N, x;
  cin >> N >> x;
  vector<long long> a(N);
  for (int i = (0); i < (N); ++i) cin >> a[i];
  sort((a).begin(), (a).end());
  vector<long long> dp(N);
  dp[0] = a[0];
  for (int i = (0); i < (N - 1); ++i) {
    dp[i + 1] = dp[i] + a[i + 1];
    for (int j = (0); j < (i + 1); ++j)
      chmin(dp[i + 1], dp[i - j] + x * (long long)(j + 1) + dp[j]);
  }
  cout << dp[N - 1] << "\n";
  return;
}
int main(void) {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(15);
  int t = 1;
  for (int _ = (0); _ < (t); ++_) {
    Main();
  }
  return 0;
}
