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;
int main() {
  long long N, x, num, ret = 0;
  vector<pair<long long, long long> > a(2000);
  vector<pair<long long, long long> > ans(2000);
  cin >> N >> x;
  for (long long i = 0; i < N; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  for (int i = 0; i < N; i++) {
    vector<pair<long long, long long> > b = a;
    for (int j = 0; j < N; j++) {
      b[j].first += ((N + (i - j)) % N) * x;
      b[j].second = ((N + (i - j)) % N);
    }
    sort(b.begin(), b.end(), greater<pair<long long, long long> >());
    ans[i].first = b[N - 1].second;
    ans[i].second = (N + i - b[N - 1].second) % N;
  }
  sort(ans.begin(), ans.end(), greater<pair<long long, long long> >());
  num = ans[0].first;
  ret = num * x;
  vector<long long> c(2000);
  for (int i = 0; i < 2 * N; i++) {
    c[i] = a[i % N].first;
  }
  for (int i = 0; i < N; i++) {
    ret += *min_element(c.begin() + i - (num + 1) + N, c.begin() + i + N);
  }
  cout << ret << endl;
  return 0;
}
