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;
struct RMQ {
  long long n;
  long long dat[2 * (4444) - 1];
  RMQ(long long n_) {
    n = 1;
    while (n < n_) n *= 2;
    for (int i = 0; i < 2 * n - 1; i++) dat[i] = 1LL << 60;
  }
  void update(int k, long long a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  long long query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return INT_MAX;
    if (a <= l && r <= b)
      return dat[k];
    else {
      long long vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
      long long vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
      return min(vl, vr);
    }
  }
};
int main() {
  long long n, x;
  cin >> n >> x;
  RMQ t(n * 2);
  for (int i = 0; i < n; i++) {
    long long a;
    cin >> a;
    t.update(i, a);
    t.update(i + n, a);
  }
  long long ans = 1LL << 60;
  for (int i = 0; i < n; i++) {
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      sum += t.query(j, j + i + 1, 0, 0, 2 * n);
    }
    ans = min(ans, sum + x * i);
  }
  cout << ans << endl;
  return 0;
}
