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;
using ll = long long;
int INF = numeric_limits<int>::max() / 2;
ll LINF = numeric_limits<ll>::max() / 2;
int MOD = 1e9 + 7;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<int> > mins(n, vector<int>(n));
  for (int i = 0; i < n; i++) {
    mins[i][0] = a[i];
  }
  for (int start = 0; start < n; start++) {
    for (int len = 1; len + start < n; len++) {
      mins[start][len] = min(mins[start][len - 1], a[start + len]);
    }
  }
  ll ans = LINF;
  for (int i = 0; i < n; i++) {
    ll temp = x * i;
    for (int j = 0; j < n; j++) {
      if (j - i >= 0)
        temp += mins[j - i][i];
      else {
        int next = j - i;
        temp += min(mins[0][j], mins[n + next][abs(next) - 1]);
      }
    }
    ans = min(ans, temp);
  }
  cout << ans << endl;
}
