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[2010];
long long maxDis[2010];
long long cnt[2010];
int main() {
  long long n, x;
  scanf("%lld%lld", &n, &x);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n; ++i) {
    long long min_j = i;
    long long min_v = a[i];
    long long min_dis = 0;
    for (long long j = 1; j <= n; ++j) {
      if (j <= i) {
        if (a[j] + (i - j) * x < min_v) {
          min_v = a[j] + (i - j) * x;
          min_j = j;
          min_dis = i - j;
        } else if (a[j] + (i - j) * x == min_v && i - j < min_dis) {
          min_dis = i - j;
          min_j = j;
        }
      } else {
        if (a[j] + (n - j + i) * x < min_v) {
          min_v = a[j] + (n - j + i) * x;
          min_j = j;
          min_dis = n - j + i;
        } else if (a[j] + (n - j + i) * x == min_v && n - j + i < min_dis) {
          min_dis = n - j + i;
          min_j = j;
        }
      }
    }
    ++cnt[min_j];
    maxDis[min_j] = max(maxDis[min_j], min_dis);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; ++i) ans += a[i] * cnt[i] + x * maxDis[i];
  printf("%lld\n", ans);
  return 0;
}
