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>
int32_t N;
int64_t x;
int32_t a[2000];
int64_t cost[2000];
int main() {
  using std::endl;
  std::cin.sync_with_stdio(false);
  std::cout.sync_with_stdio(false);
  std::cin >> N >> x;
  for (int32_t i = 0; i < N; ++i) {
    std::cin >> a[i];
  }
  for (auto& i : cost) {
    i = 1000000000000000;
  }
  int32_t shift_max = 0;
  for (int32_t i = 0; i < N; ++i) {
    for (int32_t d = 0; d < N; ++d) {
      auto b = (i - d + N) % N;
      auto tmpcost = a[b] + d * x;
      if (tmpcost < cost[i]) {
        shift_max = std::max(shift_max, d);
        cost[i] = tmpcost;
      }
    }
  }
  for (int32_t i = 0; i < N; ++i) {
    for (int32_t d = 0; d <= shift_max; ++d) {
      auto b = (i - d + N) % N;
      auto tmpcost = a[b];
      if (tmpcost < cost[i]) {
        cost[i] = tmpcost;
      }
    }
  }
  std::cout << std::accumulate(cost, cost + N, (int64_t)0) + shift_max * x
            << endl;
  return 0;
}
