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>
const long long int MOD = 1000000007;
int a[2005];
using namespace std;
int main() {
  int N, x;
  cin >> N >> x;
  for (long long int i = (long long int)(0); i < (long long int)(N); ++i) {
    cin >> a[i];
  }
  int ml = 0;
  for (long long int i = (long long int)(0); i < (long long int)(N); ++i) {
    int mI = 0;
    long long int minT = a[i];
    for (long long int l = (long long int)(0); l < (long long int)(N); ++l) {
      int j = (i + N - l) % N;
      long long int t = a[j] + x * l;
      if (t < minT) {
        minT = t;
        mI = l;
      }
    }
    ml = std::max(ml, mI);
  }
  long long int ans = x * ml;
  for (long long int i = (long long int)(0); i < (long long int)(N); ++i) {
    long long int m = a[i];
    for (long long int j = (long long int)(0); j < (long long int)(ml + 1);
         ++j) {
      int k = (i + N - j) % N;
      m = std::min(m, (long long int)a[k]);
    }
    ans += m;
  }
  cout << ans << endl;
  return 0;
}
