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>
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;
using namespace std;
int main() {
  int n;
  cin >> n;
  ;
  int x;
  cin >> x;
  ;
  vector<int> a(n);
  for (auto& youso_ : (a)) cin >> youso_;
  ;
  ll ans = 0;
  vector<int> vec;
  int ma = min_element((a).begin(), (a).end()) - a.begin();
  for (int i = (ma); i < (n); ++i) vec.push_back(a[i]);
  for (int i = 0; i < (ma); ++i) vec.push_back(a[i]);
  ans += vec[0];
  vector<int> henka(n, 0);
  vector<int> moto(n);
  for (int i = 0; i < (n); ++i) moto[i] = i;
  for (int i = (1); i < (n); ++i) {
    if (vec[i] > vec[i - 1] + x) {
      vec[i] = vec[i - 1] + x;
      moto[i] = moto[i - 1];
      henka[i] = henka[i - 1] + 1;
    }
    ans += vec[moto[i]];
  }
  ans += *max_element((henka).begin(), (henka).end()) * x;
  cout << (ans) << endl;
  ;
  return 0;
}
