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;
template <class C>
void mini(C& _a4, C _b4) {
  _a4 = min(_a4, _b4);
}
template <class C>
void maxi(C& _a4, C _b4) {
  _a4 = max(_a4, _b4);
}
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[";
  for (auto& vv : V) os << vv << ",";
  os << "]";
  return os;
}
long long n, x;
vector<long long> t;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  if (0) cout << fixed << setprecision(6);
  cin >> n >> x;
  t.resize(n);
  for (long long i = (0); i <= ((long long)(n)-1); i++) cin >> t[i];
  long long res = 1e12;
  for (long long i = (0); i <= ((long long)(n)-1); i++) {
    long long pom = i * x;
    for (long long j = (0); j <= ((long long)(n)-1); j++) pom += t[j];
    mini(res, pom);
    vector<long long> t2;
    t2.resize(n);
    for (long long j = (0); j <= ((long long)(n)-1); j++) {
      t2[j] = min(t[j], t[(j + 1) % n]);
    }
    swap(t2, t);
  }
  cout << res << "\n";
}
