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 mod = 1e9 + 7;
long long modpow(long long a, long long b, long long p) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    return modpow((a * a) % p, b / 2, p);
  } else {
    return (a * modpow(a, b - 1, p)) % p;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  ;
  for (long long i = 0; i < a.size(); i++) {
    cin >> a[i];
  };
  long long ans = LLONG_MAX;
  for (long long i = 0; i < n; i++) {
    long long time = 0;
    vector<long long> b(n);
    for (long long j = 0; j < n; j++) {
      long long Min = LLONG_MAX;
      long long kmin = 0;
      for (long long k = 0; k <= i; k++) {
        if (a[(j - k >= 0 ? j - k : n + j - k)] + k * x < Min) {
          Min = a[(j - k >= 0 ? j - k : n + j - k)] + k * x;
          kmin = k;
        }
      }
      b[j] = kmin;
    }
    for (int j = 0; j < n; j++) {
      time += a[j - b[j] >= 0 ? j - b[j] : n + j - b[j]];
    }
    sort(b.begin(), b.end());
    time += x * b[n - 1];
    ans = min(time, ans);
  }
  cout << ans << endl;
  return 0;
}
