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;
const long long int inf = 5555555555555555555;
const long long int mod = 1000000007;
const double pi = acos(-1);
const double eps = 1e-10;
long long int n = 0, m = 0, ans = 0, sum = 0, cnt = 0, tmp = 0, ma = 0, mi = 0;
string s;
bool ok, flag;
long long int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
long long int ddx[9] = {1, -1, 0, 0, 1, 1, -1, -1, 0},
              ddy[9] = {0, 0, 1, -1, 1, -1, 1, -1, 0};
long long int a[2222], b[2222];
int main(void) {
  long long int x;
  cin >> n >> x;
  for (long long int i = (long long int)(0); i < (long long int)(n); i++)
    cin >> a[i];
  for (long long int i = (long long int)(0); i < (long long int)(n); i++)
    b[i] = inf;
  for (long long int i = (long long int)(0); i < (long long int)(n); i++)
    for (long long int j = (long long int)(0); j < (long long int)(n); j++) {
      if (i == j)
        b[i] = min(b[i], a[i]);
      else {
        b[i] = min(b[i], a[j] + x);
      }
    }
  for (long long int i = (long long int)(0); i < (long long int)(n); i++)
    ans += b[i];
  cout << ans << endl;
  return 0;
}
