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 int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int M = 1000000007;
template <class T>
void debug(T a) {
  for (auto i : a) cout << i << endl;
}
signed main() {
  long long n, x;
  vector<long long> a, c;
  cin >> n >> x;
  for (int i = 0; i < (int)n; i++) {
    long long k;
    cin >> k;
    a.push_back(k);
    c.push_back(i);
  }
  auto prev = [n](int j) { return j == 0 ? n - 1 : j - 1; };
  long long ans = accumulate(begin(a), end(a), 0);
  vector<long long> b = a;
  for (int i = 0; i < (int)n; i++) {
    for (int t = 0; t < (int)n; t++) {
      c[t] = prev(c[t]);
      b[t] = min(b[t], a[c[t]]);
    }
    ans = min(ans, x * (i + 1) + accumulate(begin(b), end(b), 0));
  }
  cout << ans << endl;
}
