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;
void io_init() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
};
int main(int argc, char** argv) {
  io_init();
  int N;
  long long X;
  cin >> N >> X;
  vector<long long> A(N);
  for (long long& a : A) cin >> a;
  long long ret = 0;
  vector<bool> filled(N, false);
  for (int i = 0; i < N; i++) {
    int last = N - 1;
    while (filled[last]) last--;
    int to_fill = -1;
    long long step_cost = 1LL << 60;
    int to_move = 0;
    for (int j = 0; j < N; j++) {
      if (!filled[j]) {
        last = j;
      }
      int moves = j - last;
      if (moves < 0) moves += N;
      long long cost = moves * X + A[j];
      if (cost < step_cost) {
        step_cost = cost;
        to_fill = j;
        to_move = moves;
      }
    }
    ret += step_cost;
    if (to_move > 0) {
      vector<bool> filled2(N);
      for (int k = 0; k < N; k++) filled2[k] = filled[(k - to_move + N) % N];
      filled = filled2;
    }
    filled[to_fill] = true;
  }
  cout << ret << endl;
  return 0;
};
