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;
using int64 = int64_t;
using pii = pair<int, int>;
using pll = pair<int64_t, int64_t>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int64 n, x;
  cin >> n >> x;
  int64 a[n];
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  int64 ans = 0;
  vector<int64> v;
  for (int i = 0; i < (int)(n); ++i) {
    int64 m = 1e18;
    int64 pos = 0;
    for (int j = 0; j < (int)(n); ++j) {
      int64 temp = a[(i + j) % n] + x * j;
      if (temp < m) {
        m = temp;
        pos = j;
      }
    }
    v.push_back(pos);
    ans += m;
  }
  for (int i = 0; i < (int)(n); ++i) cout << (v[i]) << '\n';
  sort((v).begin(), (v).end());
  for (int i = n - 1; i >= 1; --i) {
    v[i] -= v[i - 1];
  }
  set<int64> s;
  for (int i = 0; i < (int)(n); ++i) {
    if (v[i] != 0) {
      s.insert(v[i]);
    }
  }
  if (s.size() == 1) {
    cout << (ans) << '\n';
    return 0;
  }
  for (auto xx : s) {
    ans -= x * xx;
  }
  cout << (ans) << '\n';
  return 0;
}
