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 int MOD = 1e9 + 7;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &i : v) is >> i;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  const string delimiter = "\n";
  for (int i = (int)(0); i < (int)(v.size()); i++) {
    os << v[i];
    if (i != v.size() - 1) os << delimiter;
  }
  return os;
}
const long long INF = 1e16;
int main() {
  int n, x;
  cin >> n >> x;
  vector<long long> a(n);
  cin >> a;
  long long ans = INF;
  vector<long long> tmp = a;
  for (int i = (int)(0); i < (int)(n); i++) {
    long long sum = 0;
    for (int j = (int)(0); j < (int)(n); j++) sum += tmp[j];
    chmin(ans, sum + i * x);
    for (int j = (int)(0); j < (int)(n); j++) {
      chmin(tmp[j], tmp[(j + n - 1) % n]);
    }
  }
  cout << ans << endl;
  return 0;
}
