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 double eps = 1e-10;
const int MOD = 1000000007;
const int INF = 1000000000;
const long long int LINF = 1ll << 50;
template <typename T>
void printv(vector<T> const& s) {
  for (int i = (0); i < (((int)(s).size())); i++) {
    cout << s[i] << " ";
  }
  cout << endl;
}
int main() {
  cin.tie(0);
  cout << setprecision(10);
  long long int n, x;
  cin >> n >> x;
  vector<long long int> a(n);
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
  }
  vector<vector<long long int> > dp(n);
  for (int i = (0); i < (n); i++) {
    dp[i].resize(n);
    dp[i][0] = a[i];
  }
  vector<vector<long long int> > c(n);
  for (int i = (0); i < (n); i++) {
    c[i].resize(n);
    c[i][0] = a[i];
  }
  for (int i = (0); i < (n); i++) {
    for (int j = (1); j < (n); j++) {
      dp[i][j] = min(dp[i][j - 1], a[(i - j + n) % n] + j * x);
      if (dp[i][j - 1] <= a[(i - j + n) % n] + j * x) {
        c[i][j] = c[i][j - 1];
      } else {
        c[i][j] = a[(i - j + n) % n];
      }
    }
  }
  long long int ans = LINF;
  for (int i = (0); i < (n); i++) {
    long long int su = 0;
    for (int j = (0); j < (n); j++) {
      su += c[j][i];
    }
    ans = min(ans, su + x * i);
  }
  cout << ans << endl;
}
