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;
using ll = long long;
void _cin() {}
template <class Head, class... Tail>
void _cin(Head&& head, Tail&&... tail) {
  cin >> head;
  _cin(forward<Tail>(tail)...);
}
void _cout() { cout << "\n"; }
template <class Head, class... Tail>
void _cout(Head&& head, Tail&&... tail) {
  cout << head;
  _cout(forward<Tail>(tail)...);
}
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
  cout << "[" << p.first << ", " << p.second << "]";
  return os;
}
int main(void) {
  int n, x;
  _cin(n, x);
  vector<int> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  ll ans = 0;
  int mmax = 0;
  for (int i = 0; i < (n); i++) {
    ll tmin = 1e18;
    int jtmin = -1;
    for (int j = 0; j < (n); j++) {
      ll t = a[(n + i - j) % n] + j * (ll)x;
      if (t <= tmin) {
        tmin = t;
        jtmin = j;
      }
    }
    ans += a[(n + i - jtmin) % n];
    mmax = max(mmax, jtmin);
  }
  _cout(ans + mmax * (ll)x);
}
