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 pii = pair<int, int>;
using vb = vector<bool>;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vvvl = vector<vvl>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vvvpii = vector<vvpii>;
template <class V>
class SegTree0 {
 private:
  static int p2(int n, int m = 1) { return m >= n ? m : p2(n, m * 2); }
  V comp(const V& l, const V& r) { return min(l, r); };
  const V def = std::numeric_limits<V>::max();

 public:
  const int n;
  vector<V> val;
  SegTree0(int _n) : n(p2(_n)), val(n * 2, def) {}
  void reinit() { val.fill(def); }
  V getval(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return def;
    if (a <= l && r <= b) return val[k];
    return comp(getval(a, b, k * 2 + 1, l, (l + r) / 2),
                getval(a, b, k * 2 + 2, (l + r) / 2, r));
  }
  V getval(int a, int b) { return getval(a, b, 0, 0, n); }
  void update(int i, V v) {
    i += n - 1;
    val[i] = v;
    while (i > 0) {
      i = (i - 1) / 2;
      val[i] = comp(val[i * 2 + 1], val[i * 2 + 2]);
    }
  }
};
void Main() {
  int N, x;
  scanf("%d%d", &N, &x);
  vi A(N);
  for (int i = int(0); i < int(N); ++i) scanf("%d", &A[i]);
  SegTree0<int> seg(N);
  for (int i = int(0); i < int(N); ++i) {
    seg.update(i, A[i]);
  }
  ll ans = 1e18;
  for (int m = int(0); m < int(N); ++m) {
    ll cand = 0;
    for (int i = int(0); i < int(N); ++i) {
      if (i >= m) {
        cand += seg.getval(i - m, i + 1);
      } else {
        cand += min(seg.getval(0, i + 1), seg.getval(N - m + i, N));
      }
    }
    ans = min(ans, cand + x * m);
  }
  (void)printf("%lld\n", ans);
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  Main();
  return 0;
}
