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 MOD = 1000000007;
long long INF = 1LL << 60;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, x;
  cin >> n >> x;
  set<int> st;
  vector<long long> a(n);
  for (long long i = (long long)(0); i < (n); i++) {
    cin >> a[i];
    if (st.count(a[i]) == 0) {
      st.insert(a[i]);
    }
  }
  long long out = INF;
  long long temp;
  for (int bound : st) {
    vector<long long> b;
    for (long long i = (long long)(0); i < (n); i++) {
      if (a[i] <= bound) {
        b.push_back(i);
      }
    }
    int k = (int)b.size();
    if (k == 1) {
      temp = a[b[0]] * n;
      temp += (n - 1) * x;
    } else {
      temp = 0;
      int maxjump = 0;
      int jump;
      for (long long i = (long long)(0); i < (k - 1); i++) {
        jump = b[i + 1] - b[i] - 1;
        maxjump = max(maxjump, jump);
      }
      jump = b[0] + n - b[k - 1] - 1;
      maxjump = max(maxjump, jump);
      for (long long i = (long long)(1); i < (k); i++) {
        jump = b[i] - b[i - 1];
        temp += jump * a[b[i]];
      }
      jump = b[0] + n - b[k - 1];
      temp += jump * a[b[0]];
      temp += (maxjump)*x;
    }
    out = min(out, temp);
  }
  cout << out;
}
