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 n, a[3000], x, b[3000], sum, minx = 99999999999999999LL;
vector<long long> vec;
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) {
    sum = 0;
    vec.clear();
    for (int j = 0; j < n; j++) {
      if (a[j] <= b[i]) vec.push_back(j);
    }
    sort(vec.begin(), vec.end());
    vec.push_back(vec[0] + n);
    for (int j = 0; j < vec.size() - 1; j++) {
      sum += (vec[(j + 1) % vec.size()] - vec[j]) * a[vec[j]];
    }
    long long mm = 0;
    for (int j = 0; j < vec.size() - 1; j++) mm = max(mm, vec[j + 1] - vec[j]);
    mm--;
    sum += mm * x;
    minx = min(minx, sum);
  }
  cout << minx << endl;
  return 0;
}
