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;
const int maxn = 2e3 + 5;
int n, X, a[maxn], cnt[maxn];
inline int rd() {
  int ret = 0, af = 1;
  char gc = getchar();
  while (gc < '0' || gc > '9') {
    if (gc == '-') af = -af;
    gc = getchar();
  }
  while (gc >= '0' && gc <= '9') ret = ret * 10 + gc - '0', gc = getchar();
  return ret * af;
}
int main() {
  n = rd();
  X = rd();
  int ti = 0;
  long long ans = 0;
  for (register int i = 1; i < n + 1; i++) a[i] = rd();
  for (register int i = 1; i < n + 1; i++) {
    long long ss = a[i];
    int pos = i;
    for (register int j = i + 1; j < n + 1; j++)
      if (ss > a[j] + (long long)(j - i) * X)
        pos = j, ss = a[j] + (long long)(j - i) * X;
    for (register int j = 1; j < i; j++)
      if (ss > a[j] + (long long)(j + n - i) * X)
        pos = j, ss = a[j] + (long long)(j + n - i) * X;
    if (pos < i)
      ti = max(ti, pos + n - i);
    else
      ti = max(ti, pos - i);
    ans += a[pos];
  }
  printf("%lld", (long long)X * ti + ans);
  return 0;
}
