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;
template <typename T>
inline void gn(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
template <typename T>
inline void gt(T x) {
  if (!x) return;
  gt(x / 10);
  putchar(x % 10 + '0');
}
const long long maxn = 4000 + 10;
struct NODE {
  double o, r, d;
};
long long t, n, m, ans;
long long MOD = 1e9 + 7, inf = 0x1f1f1f1f;
double eps = 1e-5;
long long f[maxn], a[maxn], cnt[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long i, j, k, l, ca = 0, u = 0, v = 0, d;
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> f[i], f[i + n] = f[i];
  for (i = 1 + n; i <= 2 * n; i++) {
    long long tmp = f[i];
    cnt[i]++;
    k = i;
    for (j = 1; j < n; j++) {
      if (tmp >= f[i - j] + j * m) {
        a[i] = j;
        tmp = f[i - j] + j * m;
        cnt[k]--;
        cnt[k = i - j]++;
      }
    }
    u = max(u, a[i]);
  }
  for (i = 1; i <= 2 * n; i++) ans += cnt[i] * f[i];
  ans += u * m;
  cout << ans << endl;
  return 0;
}
