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 ll = long long;
using namespace std;
const int INF = 1001001001;
int main() {
  int n, x;
  cin >> n >> x;
  vector<ll> a(n);
  for (int i = 0; i < (int)(n); i++) cin >> a.at(i);
  vector<ll> ans(n, INF);
  vector<int> ap(n, -1);
  for (int i = 0; i < (int)(n); i++) {
    ll now = a.at(i);
    if (ans.at(i) > now) {
      ans.at(i) = now;
      ap.at(i) = i;
    }
    for (int j = 0; j < (int)(n - 1); j++) {
      now += x;
      int pos = (i + j + 1) % n;
      if (ans.at(pos) > now) {
        ans.at(pos) = now;
        ap.at(pos) = i;
      }
    }
  }
  ll sum = 0;
  vector<int> co(n, 0);
  int cmx = 0;
  for (int i = 0; i < (int)(n); i++) {
    sum += a.at(ap.at(i));
    co.at(ap.at(i))++;
    cmx = max(cmx, co.at(ap.at(i)));
  }
  cerr << sum << endl;
  sum += (ll)(cmx - 1) * x;
  cout << sum << endl;
  return 0;
}
