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 int INF = 1e9, MOD = 1e9 + 7, ohara = 1e6;
const long long LINF = 1e18;
long long int n, cnt = 0, ans = INF, a[ohara], b[ohara], c, d, cmp[ohara], cmpp,
                 m, h, w, x, y, sum = 0, pos, mi[2002][2002];
int dy[] = {1, 0, -1, 0};
int dx[] = {0, 1, 0, -1};
string alph("abcdefghijklmnopqrstuvwxyz"), s;
bool fl;
struct edge {
  int to, cost;
};
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> x;
  pos = 0;
  for (long long int(i) = 0; (i) < (int)(n); (i)++) {
    cin >> a[i];
    pos += a[i];
    b[i] = INF;
  }
  for (long long int i = (n); i < (2 * n); i++) {
    a[i] = a[i - n];
  }
  for (long long int(mahou) = 0; (mahou) < (int)(n); (mahou)++) {
    for (long long int(j) = 0; (j) < (int)(n); (j)++) {
      b[j] = min(b[j], a[j + n - mahou]);
      cnt += b[j];
    }
    cnt += mahou * x;
    ans = min(ans, cnt);
    cnt = 0;
  }
  cout << (min(ans, pos)) << endl;
  return 0;
}
