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;
long long MAX = 1e18;
struct segRMQ {
  long long n;
  vector<long long> dat;
  segRMQ(long long n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.resize(2 * n - 1);
    for (int i = 0; i < 2 * n - 1; i++) dat[i] = MAX;
  }
  void update(long long 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(long long a, long long b, long long k, long long l,
                  long long r) {
    if (r <= a || b <= l) return 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;
  vector<long long> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  segRMQ segK(N);
  long long n = segK.n;
  for (int i = 0; i < N; i++) segK.update(i, a[i]);
  long long ans = MAX;
  for (long long i = 0; i < N - 1; i++) {
    long long temp = 0;
    temp += x * i;
    for (long long j = 0; j < N; j++) {
      if (j < i) {
        temp += min(segK.query(0, j + 1, 0, 0, n),
                    segK.query(n - i + j, n, 0, 0, n));
      } else
        temp += segK.query(j - i, j + 1, 0, 0, n);
    }
    ans = min(ans, temp);
  }
  cout << ans << endl;
}
