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), b(n);
  ;
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  };
  for (long long i = 0; i < n; i++) {
    long long Min = LLONG_MAX;
    long long jmin = i;
    for (long long j = 0; j < n; j++) {
      if (x * (i >= j ? i - j : n + i - j) + a[j] < Min) {
        Min = min(Min, x * (i >= j ? i - j : n + i - j) + a[j]);
        jmin = j;
      }
    }
    b[i] = (jmin - i < 0 ? jmin - i : jmin - n - i);
  }
  vector<bool> use(n);
  for (long long i = 0; i < n; i++) {
    use[i] = false;
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += a[(i + b[i] >= 0 ? i + b[i] : n + i + b[i])];
  }
  sort(b.begin(), b.end());
  cout << ans - x * b[0] << endl;
  return 0;
}
