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 N = (int)(2e3 + 5);
const int LOG = (int)(20);
long long int n, cs, ar[N], mini[N][N], tree[N], ans = INT_MAX;
void build(int ind, int x, int y) {
  if (x == y) {
    tree[ind] = ar[x];
    return;
  }
  build((ind * 2), x, ((x + y) / 2));
  build((ind * 2 + 1), ((x + y) / 2) + 1, y);
  tree[ind] = min(tree[(ind * 2)], tree[(ind * 2 + 1)]);
}
long long int get(int ind, int x, int y, int a, int b) {
  if (a > b) return min(get(ind, x, y, 1, b), get(ind, x, y, a, n));
  if (y < a || b < x) return INT_MAX;
  if (a <= x && y <= b) return tree[ind];
  return min(get((ind * 2), x, ((x + y) / 2), a, b),
             get((ind * 2 + 1), ((x + y) / 2) + 1, y, a, b));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> cs;
  for (int i = 1; i <= n; i++) cin >> ar[i];
  build(1, 1, n);
  for (int k = 0; k <= n; k++) {
    long long int t = k * cs;
    for (int i = 1; i <= n; i++) {
      int l = ((i - k) < 1) ? i - k + n : i - k;
      int r = i;
      t += get(1, 1, n, l, r);
    }
    ans = min(ans, t);
  }
  cout << ans;
}
