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() {
  unsigned long long int n, x;
  cin >> n >> x;
  vector<unsigned long long int> slime;
  unsigned long long int tmin = 1000000007, j = -1;
  for (int i = 0; i < n; i++) {
    unsigned long long int tmp;
    cin >> tmp;
    if (tmin > tmp) {
      tmin = tmp;
      j = i;
    }
    slime.push_back(tmp);
  }
  unsigned long long int ans = 0, max_ij = 0;
  for (int i = 0; i < n; i++) {
    unsigned long long int tmp;
    if (i == j) {
      tmp = tmin;
    } else if (i > j) {
      tmp = min(tmin + x * (i - j), slime[i]);
      if (i - j <= max_ij) {
        tmp -= x * (i - j);
      } else {
        max_ij = i - j;
      }
    } else {
      tmp = min(tmin + x * (i - j + n), slime[i]);
      if (i - j + n <= max_ij) {
        tmp -= x * (i - j + n);
      } else {
        max_ij = i - j + n;
      }
    }
    ans += tmp;
  }
  cout << ans;
}
