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 mod = (int)1e9 + 7;
struct Segtree {
  vector<int> seg;
  int nbn;
  int INF;
  void init(int n) {
    for (nbn = 1; nbn < n + 2; nbn <<= 1)
      ;
    INF = 1 << 29;
    seg.assign(nbn * 2, INF);
  }
  int comp(int l, int r) { return min(l, r); }
  void update(int k, int x) {
    k += nbn - 1;
    seg[k] = x;
    while (k > 0) {
      k = (k - 1) / 2;
      seg[k] = comp(seg[k * 2 + 1], seg[k * 2 + 2]);
    }
  }
  int query2(int a, int b, int l, int r, int k) {
    if (b <= l || r <= a) return INF;
    if (a <= l && r <= b) return seg[k];
    return comp(query2(a, b, l, (l + r) / 2, k * 2 + 1),
                query2(a, b, (l + r) / 2, r, k * 2 + 2));
  }
  int query(int a, int b) { return query2(a, b, 0, nbn, 0); }
  int get(int a) { return seg[nbn + a - 1]; }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  vector<int> a(n);
  for (int(i) = 0; (i) < (n); (i)++) cin >> a[i];
  Segtree sga;
  sga.init(2 * n + 10);
  for (int(i) = 0; (i) < (n); (i)++) {
    sga.update(i, a[i]);
    sga.update(i + n, a[i]);
  }
  long long ans = accumulate((a).begin(), (a).end(), 0LL);
  for (int(i) = (1); (i) < (n); (i)++) {
    long long w = 0;
    for (int(j) = 0; (j) < (n); (j)++) w += sga.query(j + n - i, j + n + 1);
    ans = min(ans, w + (long long)x * i);
  }
  cout << ans << endl;
  return 0;
}
