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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (T &x : vec) is >> x;
  return is;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int n, x;
  cin >> n >> x;
  vector<long long int> a(n);
  cin >> a;
  long long int ans = 0;
  int max_shift = 0;
  for (int i = 0; i < n; i++) {
    long long int minv = (1LL << 50);
    long long int vv = (1LL << 50);
    for (int j = 0; j < n; j++) {
      long long int v = a[(i - j + n) % n] + x * j;
      if (minv > v) {
        vv = a[(i - j + n) % n];
        minv = v;
        max_shift = max(max_shift, j);
      }
    }
    ans += vv;
  }
  cout << ans + x * max_shift << endl;
  return 0;
}
