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;
int N, x;
long long a[2016];
long long mini[2016];
bool fin[2016];
void compute() {
  for (int i = 0; i < N; ++i) mini[i] = a[i];
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j)
      if (i != j) {
        if (i < j) {
          mini[j] = min(mini[j], a[i] * ((j - i) + 1LL));
        } else {
          mini[j] = min(mini[j], a[i] * (j - i + N));
        }
      }
  int remain = N;
  long long answer = 0;
  deque<int> sp;
  for (int i = 0; i < N; ++i) {
    if (mini[i] == a[i]) {
      fin[i] = true;
      --remain;
      answer += a[i];
      sp.push_back(i);
    }
  }
  for (int i = 0; i < (int)sp.size(); ++i) {
    long long step = 0;
    int cur = (sp[i] + 1) % N;
    while (!fin[cur]) {
      (cur += 1) %= N;
      ++step;
    }
    answer += (mini[sp[i]] * step + x * step);
  }
  cout << answer << endl;
}
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; ++i) cin >> a[i];
  compute();
  return 0;
}
