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 solve(long long N, long long x, vector<long long> a) {
  long long ans = 0;
  int no_move = 0;
  int move_max = 0;
  for (int i = (0); i < (N); ++i) {
    long long min_cost = 1e9;
    int moves = N * 2;
    int idd = 0;
    for (int j = (0); j < (N); ++j) {
      if (i == j) continue;
      int diff = i - j;
      if (diff < 0) diff += N;
      long long cost = a[j] + diff * x;
      if (min_cost == cost) {
        moves = min(moves, diff);
        if (moves == diff) idd = j;
      } else {
        min_cost = min(min_cost, cost);
        if (min_cost == cost) {
          moves = diff;
          idd = j;
        }
      }
    }
    if (min_cost >= a[i]) {
      no_move++;
      ans += a[i];
    } else {
      ans += a[idd];
      move_max = max(moves, move_max);
    }
  }
  ans += x * move_max;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  long long N;
  long long x;
  cin >> N;
  vector<long long> a(N - 1 + 1);
  cin >> x;
  for (int i = 0; i <= N - 1; i++) {
    cin >> a[i];
  }
  solve(N, x, a);
  return 0;
}
