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 = int, class U = int>
class SlideMin {
  struct Pair {
    T index;
    U value;
    Pair(){};
    Pair(T index, U value) : index(index), value(value){};
  };

 private:
  deque<Pair> que;
  T length;

 public:
  SlideMin(T length) { this->length = length; }
  Pair top() { return this->que.front(); }
  void update(T index, U value) {
    while (!this->que.empty() && this->que.back().value >= value) {
      this->que.pop_back();
    }
    this->que.emplace_back(index, value);
    if (this->que.front().index == index - this->length) this->que.pop_front();
  }
};
int N, X;
vector<int> A;
vector<int> B;
long long sol(int n) {
  SlideMin<int, long long> que(n + 1);
  vector<long long> ret(N, 10000000000);
  for (int i = 0LL; i < int(2 * N); i++) {
    int b = B[i];
    que.update(i, b);
    ret[i % N] = min(ret[i % N], que.top().value);
  }
  long long res = 0;
  for (long long n : ret) {
    res += n;
  }
  return res + n * X;
}
int main(void) {
  cin >> N >> X;
  A.resize(N);
  B.resize(N * 2);
  for (int i = 0LL; i < int(N); i++) {
    cin >> A[i];
    B[i] = A[i];
    B[i + N] = A[i];
  };
  long long ans = 0;
  for (int a : A) {
    ans += a;
  }
  for (int i = 0LL; i < int(N); i++) {
    ans = min(ans, sol(i));
  }
  cout << ans << endl;
  return 0;
}
