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 ll = long long;
using P = pair<int, int>;
const ll MOD = 1000000007;
const ll INF = 1LL << 60;
template <typename T>
class SegmentTree {
 private:
  int n;
  vector<T> dat;

 public:
  void init(int n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.assign(2 * n - 1, INF);
  }
  void init(vector<T>& value) {
    int n_ = value.size();
    n = 1;
    while (n < n_) n *= 2;
    dat.resize(2 * n - 1);
    for (int i = 0; i < n_; i++) {
      dat[i + n - 1] = value[i];
    }
    for (int i = n_ + n - 1; i < 2 * n - 1; i++) {
      dat[i] = INF;
    }
    for (int k = n - 2; k >= 0; k--) {
      dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  void update(int k, int a) {
    k += n - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  T query(int a, int b) { return _query(a, b, 0, 0, n - 1); }
  T _query(int a, int b, int k, int l, int r) {
    if (r < a || b < l) return INF;
    if (a <= l && r <= b) {
      return dat[k];
    } else {
      T vl = _query(a, b, k * 2 + 1, l, (l + r) / 2);
      T vr = _query(a, b, k * 2 + 2, (l + r) / 2 + 1, r);
      return min(vl, vr);
    }
  }
  T get(int k) { return dat[k + n - 1]; }
};
int main() {
  int N, x;
  cin >> N >> x;
  SegmentTree<ll> seg;
  vector<ll> a(2 * N);
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    a[i + N] = a[i];
  }
  seg.init(a);
  ll ans = INF;
  for (int m = 0; m < N; m++) {
    ll ans_m = x * m;
    for (int i = 0; i < N; i++) {
      ans_m += seg.query(N + i - m, N + i);
    }
    ans = min(ans, ans_m);
  }
  cout << ans << endl;
  return 0;
}
