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;
void CINT() {}
template <class Head, class... Tail>
void CINT(Head&& head, Tail&&... tail) {
  cin >> head;
  CINT(move(tail)...);
}
const long long LINF = 1e18 + 1;
const int MOD = 1e9 + 7;
const int MAX_N = 1e5 + 1;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  CINT(N);
  long long x;
  CINT(x);
  vector<long long> a(N);
  long long mnNum = LINF;
  int mn = 0;
  for (int i = (int)0; i < (int)N; ++i) {
    cin >> a[i];
    if (a[i] < mnNum) {
      mn = i;
      mnNum = a[i];
    }
  }
  long long ans = LINF;
  for (int h = (int)0; h < (int)N; ++h) {
    long long tmp = 0ll;
    int xNum = 0;
    for (int i = (int)0; i < (int)N; ++i) {
      int j = mn;
      if (h == j) {
        tmp += a[h];
      } else {
        if (a[h] < ((h - j + N) % N) * x) {
          tmp += min(a[h], a[j] + ((h - j + N) % N) * x);
        } else {
          xNum = max(xNum, (h - j + N) % N);
          tmp += a[j];
        }
      }
    }
    tmp += xNum * x;
    ans = min(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}
