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;
template <class T>
class DequeSlide {
  T *data;
  int *deq;
  int n, l, s, t;
  int cur;

 public:
  DequeSlide(int n, int l) {
    data = new T[n];
    deq = new int[n];
    this->n = n;
    this->l = l;
    s = t = cur = 0;
  }
  void push(int x) {
    if (s < t && deq[s] == cur - l) s++;
    data[cur++] = x;
    while (s < t && data[deq[t - 1]] >= x) t--;
    deq[t++] = cur - 1;
  }
  T get() {
    if (s == t) return 100000000000000007LL;
    return data[deq[s]];
  }
  void debug() {
    cout << s << " " << t << " [";
    for (int i = s; i <= t - 1; ++i) {
      cout << " " << data[deq[i]];
    }
    cout << " ]" << endl;
  }
};
int main() {
  int n, x;
  cin >> n >> x;
  long long A[2000], ans = 100000000000000007LL;
  for (int i = 0; i < n; ++i) cin >> A[i];
  for (int l = 1; l <= n; ++l) {
    DequeSlide<long long> deq(2 * n, l);
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      deq.push(A[i]);
    }
    for (int i = n; i <= 2 * n - 1; ++i) {
      deq.push(A[i % n]);
      res += deq.get();
    }
    ans = (ans > (res + x * (l - 1)) ? (res + x * (l - 1)) : ans);
  }
  cout << ans << endl;
  return 0;
}
