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;
vector<long long> divisors(long long n) {
  vector<long long> res;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      res.push_back(i);
      if (i != n / i) res.push_back(n / i);
    }
  }
  sort(res.begin(), res.end());
  return res;
}
int main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 1e18;
  for (int d : divisors(n)) {
    for (int i = 0; i < d; i++) {
      long long sum = 0;
      for (int j = i; j < n; j += d) {
        sum += a[j];
      }
      ans = min(ans, sum * d + x * (d - 1));
    }
  }
  cout << ans << endl;
}
