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;
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.push_back(x);
  return v;
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << "[DEBUG] " << it->substr((*it)[0] == ' ', it->length()) << " = " << a
       << '\n';
  err(++it, args...);
}
const int MOD = 1000000007;
template <class T, class U>
inline T tmin(T a, U b) {
  return (a < b) ? a : b;
}
template <class T, class U>
inline T tmax(T a, U b) {
  return (a > b) ? a : b;
}
template <class T, class U>
inline void amax(T &a, U b) {
  if (b > a) a = b;
}
template <class T, class U>
inline void amin(T &a, U b) {
  if (b < a) a = b;
}
template <class T>
inline T tabs(T a) {
  return (a > 0) ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  while (b != 0) {
    T c = a;
    a = b;
    b = c % b;
  }
  return a;
}
const int N = 2001;
bool vis[N];
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  long long ans = 0;
  int cnt = 0;
  memset(vis, 0, sizeof(vis));
  const int pos = min_element(a.begin(), a.end()) - a.begin();
  const int mini = a[pos];
  vis[pos] = 1;
  int cur = pos;
  while (cnt < n) {
    while (cnt < n && a[cur] < mini + x) {
      ans += a[cur];
      vis[cur] = 1, cnt++;
      cur = (cur + 1) % n;
    }
    if (cnt < n) {
      ans += x + mini;
      cur = (cur + 1) % n, cnt++;
    }
  }
  cout << ans << endl;
  return 0;
}
