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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, x, a[2005 * 2], rk[2005];
bool used[2005];
bool cmp(int x, int y) { return a[x] < a[y]; }
int main() {
  n = read(), x = read();
  register int i, j;
  for (i = 1; i <= n; i++) a[i] = a[i + n] = read(), rk[i] = i;
  sort(rk + 1, rk + n + 1, cmp);
  long long ans = 0;
  for (i = 1; i <= n; ++i) {
    if (used[rk[i]]) continue;
    used[rk[i]] = 1;
    ans += a[rk[i]];
    int plus = 1;
    for (j = rk[i] + 1; j < rk[i] + n; j++) {
      if (a[j] < a[rk[i]] + x * plus) break;
      if (!used[j % n ? j % n : n])
        ans += a[rk[i]] + x * plus, used[j % n ? j % n : n] = 1, plus = 1;
      else
        plus++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
