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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  long long int n, x;
  long long int a[2000];
  long long int k[2000];
  long long int maxkaisu = 0;
  long long int sum = 0;
  cin >> n >> x;
  for (int i = 0; i < (int)(n); i++) cin >> a[i];
  for (int i = 0; i < (int)(n); i++) {
    vector<pair<long long int, long long int> > p;
    for (int j = 0; j < (int)(n); j++) {
      if (i == j) continue;
      int t;
      t = i - j;
      if (t < 0) t += n;
      p.push_back(make_pair(a[j] + x * t, t));
    }
    sort(p.begin(), p.end());
    long long int kaisu;
    if (p[0].first > a[i]) {
      kaisu = 0;
      k[i] = a[i];
    } else {
      kaisu = p[0].second;
      k[i] = p[0].first - x * p[0].second;
    }
    maxkaisu = max(kaisu, maxkaisu);
  }
  for (int i = 0; i < (int)(n); i++) sum += k[i];
  cout << sum + x * maxkaisu << endl;
}
