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 void read(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
inline void writeln(int x) {
  char buffor[21];
  register int i = 0;
  int neg = 0;
  if (x < 0) {
    neg = 1;
    x = -x;
  }
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  if (neg) putchar('-');
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
const int N = 2010;
int n;
long long a[N], x, ok[N], id[N];
void input() {
  cin >> n >> x;
  for (int i = 0; i <= n - 1; i++) cin >> a[i];
}
bool bya(int i, int j) { return a[i] < a[j]; }
void solve() {
  for (int i = 0; i <= n - 1; i++) id[i] = i;
  sort(id + 0, id + 0 + n, bya);
  long long res = 1e15;
  for (int dich = 0; dich <= n - 1; dich++) {
    fill(ok + 0, ok + 0 + n, 0);
    long long mini_res = 0;
    for (int i = 0; i <= n - 1; i++) {
      int cur = id[i];
      for (int j = 0; j <= dich; j++)
        if (ok[(cur + j) % n] == 0) {
          ok[(cur + j) % n] = 1;
          mini_res += a[cur];
        }
    }
    res = min(res, mini_res + x * dich);
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
