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;
using lint = long long int;
using P = pair<lint, lint>;
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) { cout << (expr ? "Yes" : "No") << "\n"; }
struct Segment_Min {
 private:
  int n;
  vector<int> node;

 public:
  Segment_Min(vector<int> v) {
    int sz = v.size();
    n = 1;
    while (n < sz) n *= 2;
    node.resize(2 * n - 1, INF);
    for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      node[i] = min(node[2 * i + 1], node[2 * i + 2]);
  }
  void update(int x, int val) {
    x += (n - 1);
    node[x] = val;
    while (x > 0) {
      x = (x - 1) / 2;
      node[x] = min(node[2 * x + 1], node[2 * x + 2]);
    }
  }
  int getmin(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    if (r <= a || b <= l) return INF;
    if (a <= l && r <= b) return node[k];
    int vl = getmin(a, b, 2 * k + 1, l, (l + r) / 2);
    int vr = getmin(a, b, 2 * k + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int N, x;
  cin >> N >> x;
  vector<int> a(2 * N);
  for (int i = (0), i_end_ = (N); i < i_end_; i++) cin >> a[i];
  for (int i = (0), i_end_ = (N); i < i_end_; i++) a[N + i] = a[i];
  Segment_Min seg(a);
  lint ans = 1e18;
  for (int k = (0), k_end_ = (N); k < k_end_; k++) {
    lint tmp = k * x;
    for (int i = (0), i_end_ = (N); i < i_end_; i++)
      tmp += seg.getmin(N + i - k, N + i + 1);
    if (tmp < ans) ans = tmp;
  }
  cout << ans << endl;
}
