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>
const long long MOD = 1e9 + 7;
const long long MAXN = 1e6 + 1;
using namespace std;
long long readInt() {
  bool minus1 = false;
  long long result = 0;
  char ch;
  ch = getchar();
  while (true) {
    if (ch == '-') break;
    if (ch >= '0' && ch <= '9') break;
    ch = getchar();
  }
  if (ch == '-')
    minus1 = true;
  else
    result = ch - '0';
  while (true) {
    ch = getchar();
    if (ch < '0' || ch > '9') break;
    result = result * 10 + (ch - '0');
  }
  if (minus1)
    return -result;
  else
    return result;
}
long long a[MAXN];
long long ans[MAXN];
long long kek[MAXN];
bool used[MAXN];
int main() {
  int n = readInt(), x = readInt();
  for (int i = 1; i <= n; i++) {
    a[i] = readInt();
  }
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    long long mn = 10 * MOD;
    int tmp = i;
    int mx_tmp = 0;
    for (int j = 1; j <= n; j++) {
      if (mn > (j - 1) * x + a[tmp]) {
        mn = min(mn, (j - 1) * x + a[tmp]);
        mx_tmp = j - 1;
        ans[i] = a[tmp];
      }
      if (tmp == 1) {
        tmp = n;
      } else {
        tmp--;
      }
    }
    mx = max(mx, mx_tmp);
  }
  long long res1 = 0;
  for (int i = 1; i <= n; i++) {
    res1 += ans[i];
  }
  cout << res1 + (mx * x);
  return 0;
}
