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 long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long f[4010][4010], a[4010];
long long n, x, ans = 1e18;
int main() {
  n = read(), x = read();
  for (register long long i = 1; i <= n; i++) a[i] = read(), a[i + n] = a[i];
  for (register int i = 0; i <= (n << 1); i++) {
    for (register int j = 0; j <= (n << 1); j++) f[i][j] = 1e18;
  }
  for (register int i = 1; i <= (n << 1); i++) {
    for (register int j = i; j >= 1; j--) {
      if (i == j) {
        f[i][j] = min(a[i], f[i - 1][i - 1] + x);
        continue;
      } else
        f[j][i] = min(f[j - 1][i - 1] + x,
                      min(f[j][i - 1] + a[i], f[j + 1][i] + a[j]));
    }
  }
  for (register int i = 1; i <= n; i++) ans = min(ans, f[i][i + n - 1]);
  printf("%lld\n", ans);
  return 0;
}
