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 INF = 1e9;
const int MOD = 1e9 + 7;
const long long LLINF = 1e18;
static const double pi = 3.141592653589793;
int N, X;
long long A[4009];
long long calc(int K) {
  map<long long, int> dic;
  for (int i = 0; i < K; i++) dic[A[i]]++;
  long long res = 0;
  for (int i = 0; i < N; i++) {
    int idx = i + K;
    dic[A[idx]]++;
    auto itr = dic.begin();
    res += itr->first;
    if (--dic[A[i]] <= 0) dic.erase(A[i]);
  }
  res += X * K;
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> X;
  for (int i = 0; i < N; i++) cin >> A[i];
  for (int i = 0; i < N; i++) A[i + N] = A[i];
  long long ans = 0;
  for (int i = 0; i < N; i++) ans += A[i];
  for (int i = 1; i <= N - 1; i++) ans = min(ans, calc(i));
  ;
  cout << ans << endl;
}
