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 main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> times(n, 0);
  vector<int> start(n, 0);
  for (int i = 0; i < n; ++i) {
    cin >> times[i];
    start[i] = i;
  }
  long long switch_time = 0;
  int last_switch = 1;
  for (int i = 1; i < n; ++i) {
    bool is_switched = false;
    for (int j = 0; j < n; ++j) {
      int prev = (start[j] + n - i) % n;
      if (times[prev] + (last_switch * x) < times[start[j]]) {
        switch_time += (last_switch * x);
        last_switch = 1;
        is_switched = true;
        break;
      }
    }
    if (is_switched) {
      for (int j = 0; j < n; ++j) {
        int prev = (start[j] + n - i) % n;
        if (times[prev] < times[start[j]]) {
          start[j] = prev;
        }
      }
    } else {
      ++last_switch;
    }
  }
  long long min_time = 0;
  for (int i = 0; i < n; ++i) {
    min_time += times[start[i]];
  }
  cout << min_time + switch_time << endl;
  return 0;
}
