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 double pi = acos(-1.0);
const double eps = 1e-9;
using namespace std;
const int N = 2e5 + 1;
long long dp[N];
void solve() {
  int n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  long long ans = 1e18;
  for (int from = 0; from < n; from++) {
    for (int i = 0; i < n; i++) dp[i] = 0;
    int id = (from + 1) % n;
    dp[from] = a[from];
    while (id != from) {
      dp[id] =
          min(a[id] + dp[(id - 1 + n) % n], dp[(id - 1 + n) % n] + m + a[from]);
      id++;
      id %= n;
    }
    ans = min(ans, dp[(from - 1 + n) % n]);
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(NULL), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(10);
  int step;
  step = 1;
  for (int i = 1; i <= step; i++) {
    solve();
  }
}
