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 MAX_N = 2e3 + 1;
int N, X;
long long A[MAX_N];
long long P[MAX_N];
const long long Inf = 1e15;
int f(int from, int to) { return from <= to ? to - from : N - (from - to); }
void solve() {
  long long res = Inf;
  for (int i = 0; i < (N); ++i) P[i] = i;
  for (int d = 0; d < (N); ++d) {
    long long tot = 0;
    for (int i = 0; i < (N); ++i) {
      int dold = f(P[i], i);
      int p = (i - d + N) % N;
      long long a = 1LL * X * dold + A[P[i]], b = 1LL * X * d + A[p];
      if (a > b) P[i] = p;
      tot += A[P[i]];
    }
    res = min(res, tot + X * d);
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> X;
  for (int i = 0; i < (N); ++i) cin >> A[i];
  solve();
  return 0;
}
