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;
const double PI = 2 * acos(.0);
const int INF = 0x3f3f3f3f;
const long long LLINF = 1000000000000000005LL;
;
const long long MOD = (long long)(1e9) + 7;
void _mod(long long& x) { x = ((((x) % (MOD)) + (MOD)) % (MOD)); }
const double EPS = 1e-10;
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T& val) {
  std::fill((T*)array, (T*)(array + N), val);
}
class BColorfulSlimes {
 public:
  int N, x;
  void solve(std::istream& cin, std::ostream& cout) {
    cin >> N >> x;
    long long mov[N], A[N];
    memset(mov, 0, sizeof(mov));
    memset(A, 0, sizeof(A));
    for (__typeof(N) i = 0; i < (N); i += 1) cin >> A[i];
    for (__typeof(N) n = 0; n < (N); n += 1) {
      long long mi = 1e18, mv;
      for (__typeof(N) m = 0; m < (N); m += 1)
        if (mi > m * x + A[(n + m) % N])
          mi = m * x + A[(n + m) % N], mov[n] = m;
    }
    long long sum = 0, mx = -1;
    for (__typeof(N) i = 0; i < (N); i += 1)
      sum += A[(mov[i] + i) % N], mx = max(mx, mov[i]);
    sum += mx * x;
    cout << sum << endl;
  }
};
int main() {
  BColorfulSlimes solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
