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;
long long int a[2005];
long long int price[2005];
long long int coun[2005];
int main() {
  long long int N, x, maxi = 0, Ans = 0;
  scanf("%lld %lld", &N, &x);
  for (long long int i = 0; i < N; ++i) {
    scanf("%lld", a + i);
    price[i] = a[i];
    coun[i] = i;
  }
  long long int temp;
  long long int tempmax;
  for (long long int i = 0; i < N; ++i) {
    tempmax = 0;
    for (long long int j = 0; j < N; ++j) {
      temp = a[j] + ((i - j + N) % N) * x;
      if (price[i] > temp) {
        price[i] = temp;
        tempmax = (i - j + N) % N;
        coun[i] = j;
      } else if (price[i] == temp) {
        if (tempmax < (i - j + N) % N) {
          tempmax = (i - j + N) % N;
          coun[i] = j;
        }
      }
    }
    maxi = max(maxi, tempmax);
  }
  Ans = maxi * x;
  for (long long int i = 0; i < N; ++i) Ans += a[coun[i]];
  printf("%lld\n", Ans);
}
