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 int INF = 1e8;
using namespace std;
template <typename T>
void dpite(T a, T b) {
  return;
}
template <typename T>
void pite(T a, T b) {
  for (T ite = a; ite != b; ite++) cout << (ite == a ? "" : " ") << *ite;
  cout << '\n';
}
template <typename T>
bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
void ios_init() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
int main() {
  ios_init();
  int n;
  while (cin >> n) {
    long long int x;
    cin >> x;
    vector<long long int> a(n);
    for (int i = (0); i < (n); ++i) cin >> a[i];
    long long int ans = accumulate((a).begin(), (a).end(), 0);
    vector<long long int> mm = a;
    for (int k = (0); k < (n); ++k) {
      for (int i = (0); i < (n); ++i) {
        mm[i] = min(mm[i], a[(i - k + n) % n]);
      }
      ans = min(ans, accumulate((mm).begin(), (mm).end(), 0) + k * x);
    }
    cout << ans << '\n';
  }
  return 0;
}
