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(void) {
  int n, x, flag = 0, check = 0, capture, minkey = 0;
  long long ans = 0, count = 0;
  cin >> n >> x;
  vector<int> a(n), slime(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < a[minkey]) minkey = i;
  }
  while (flag == 0) {
    capture = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] <= x && slime[i] == 0) {
        ans += a[i];
        slime[i] = 1;
        capture = 1;
      }
    }
    if (!capture) {
      ans += a[minkey];
      slime[minkey] = 1;
    }
    flag = 1;
    check = slime[n - 1];
    slime[n - 1] = 0;
    for (int i = n - 1; i > 0; i--) {
      if (slime[i - 1] == 1) {
        slime[i] = 1;
        slime[i - 1] = 0;
      }
    }
    slime[0] = check;
    for (int i = 0; i < n; i++) {
      if (slime[i] == 0) flag = 0;
    }
    if (flag == 0) ans += x;
  }
  cout << ans << endl;
  return 0;
}
