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;
constexpr long long INF = 1e18;
constexpr long long MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int dx[] = {0, 0, 1, -1};
constexpr int dy[] = {1, -1, 0, 0};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  long long x, a[2123];
  pair<long long, int> a_[2123];
  cin >> N >> x;
  for (int i = (0); i < (N); i++) {
    cin >> a[i];
    a_[i].first = a[i];
    a_[i].second = i + 1;
  }
  sort(a_, a_ + N);
  bool used[2123] = {false};
  int usedNum = 0;
  int i = 0;
  long long res = 0;
  long long res_x = 0;
  while (usedNum < N) {
    if (used[a_[i].second]) {
      i++;
      continue;
    }
    used[a_[i].second] = true;
    usedNum++;
    long long num = a_[i].first;
    int j = a_[i].second - 1;
    int cnt = 0;
    res += a_[i].first;
    while (true) {
      if (usedNum >= N) break;
      num += x + a_[i].first;
      j++;
      if (j == N) j = 0;
      if (used[j + 1]) break;
      if (num > a[j]) break;
      usedNum++;
      used[j + 1] = true;
      res += a_[i].first;
      cnt++;
    }
    res_x = max(res_x, (long long)cnt * x);
    i++;
  }
  cout << res + res_x << endl;
  return 0;
}
