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;
const static int MOD = 1000000000 + 7;
const static long long int INF = 1LL << 60;
const static int dy[] = {0, 1, 0, -1};
const static int dx[] = {1, 0, -1, 0};
signed main(int argc, char *argv[]) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  long long int x;
  cin >> N >> x;
  vector<long long int> slime(N);
  for (int i = 0; i < N; ++i) {
    cin >> slime[i];
  }
  priority_queue<long long int, vector<long long int>, greater<long long int> >
      PQ;
  for (int i = 0; i < N; ++i) {
    long long int ans = 0;
    for (int j = 0; j < N; ++j) {
      if (i == j)
        ans += slime[i];
      else
        ans += min(slime[j], x + slime[i]);
    }
    PQ.push(ans);
  }
  cout << PQ.top() << endl;
  return 0;
}
