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>
long long n, a[4001], x, b[2001], c[2001];
long long d[2001];
int absl(int a) {
  if (a < 0) {
    return -a;
  } else {
    return a;
  }
}
int main(void) {
  int i, j, max = 0;
  long long ans = 0;
  scanf("%lld %lld", &n, &x);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    a[i + n] = a[i];
  }
  for (i = 1; i <= n; i++) {
    d[i] = a[i];
    b[i] = a[i];
    for (j = i; j <= n + n; j++) {
      if (d[i] > a[j] + x * absl((i + n) - j)) {
        b[i] = a[j];
        c[i] = absl((i + n) - j);
        if (max < absl((i + n) - j)) {
          max = absl((i + n) - j);
        }
        d[i] = b[i] * c[i];
      }
    }
    ans += b[i];
  }
  ans += x * max;
  printf("%lld\n", ans);
  return 0;
}
