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 a[2123];
long long b[2123];
long long x;
long long m[2123];
int main() {
  int N;
  long long ans;
  cin >> N >> x;
  int i;
  int j;
  for (i = 0; i < N; i++) {
    scanf("%d", a + i);
    b[i] = a[i];
    m[i] = x * i;
  }
  int big[N];
  int MAX = 0;
  fill(big, big + N, 0);
  for (i = 0; i < N; i++) {
    for (j = 0; j < N; j++) {
      if (i + j < N) {
        if (a[i + j] > a[i] + m[j]) {
          big[i + j] = j;
        }
        a[i + j] = min(a[i + j], (a[i] + m[j]));
      } else {
        if (a[i + j - N] > a[i] + m[j]) {
          big[i + j - N] = j;
        }
        a[i + j - N] = min(a[i + j - N], (a[i] + m[j]));
      }
    }
  }
  ans = 0;
  for (i = 0; i < N; i++) {
    MAX = max(MAX, big[i]);
  }
  for (i = 0; i < N; i++) {
    if (big[i]) {
      if (big[i] == MAX) {
        MAX = N;
      } else {
        a[i] -= big[i] * x;
      }
    }
    ans += a[i];
  }
  cout << ans << endl;
  return 0;
}
