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;
long long inv_mod(long long a, long long m) {
  long long b, x, u, q, tmp;
  b = m;
  x = 1;
  u = 0;
  while (b > 0) {
    q = a / b;
    tmp = u;
    u = x - q * u;
    x = tmp;
    tmp = b;
    b = a - q * b;
    a = tmp;
  }
  return (x < 0) ? m + x : x;
}
signed main() {
  long long N, x;
  vector<long long> a;
  cin >> N >> x;
  for (int i = 0; i < (N); i++) {
    long long tmp;
    cin >> tmp;
    a.push_back(tmp);
  }
  vector<vector<pair<long long, long long>>> arr;
  long long m = 0;
  arr.resize(N);
  for (int i = 0; i < (N); i++) {
    for (int j = 0; j < (N); j++) {
      arr[i].push_back(pair<long long, long long>(a[(i + j) % N] + j * x, j));
    }
    sort((arr[i]).begin(), (arr[i]).end(),
         [](pair<long long, long long> p, pair<long long, long long> q) {
           return p.first < q.first;
         });
    m = max(m, arr[i][0].second);
  }
  long long ans = 0;
  for (int i = 0; i < (N); i++) ans += (arr[i][0].first - arr[i][0].second * x);
  cout << ans + m * x << endl;
  return 0;
}
