Problem:
Score : 600 points

Problem Statement

Given an undirected tree, let the distance between vertices u and v be the
number of edges on the simple path from u to v.
The diameter of a tree is the maximum among the distances between any two vertices.
We will call a tree good if and only if its diameter is at most K.

You are given an undirected tree with N vertices numbered 1 through N.
For each i (1≦i≦N-1), there is an edge connecting vertices A_i and B_i.

You want to remove zero or more vertices from the tree, so that the resulting tree is good.
When a vertex is removed, all incident edges will also be removed.
The resulting graph must be connected.

Find the minimum number of vertices that you need to remove in order to produce a good tree.

Constraints

2≦N≦2000

1≦K≦N-1

1≦A_i≦N, 1≦B_i≦N

The graph defined by A_i and B_i is a tree.

Input

The input is given from Standard Input in the following format:

N K
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}

Output

Print the minimum number of vertices that you need to remove in order to produce a good tree.

Sample Input 1

6 2
1 2
3 2
4 2
1 6
5 6

Sample Output 1

2

The tree is shown below. Removing vertices 5 and 6 will result in a good tree with the diameter of 2.

Sample Input 2

6 5
1 2
3 2
4 2
1 6
5 6

Sample Output 2

0

Since the given tree is already good, you do not need to remove any vertex.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define BIT(n) (1LL << (n))
#define BITF(n, i) (((n) >> i) & 1)
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPB(i, n) for (int i = 0; i < BIT(n); i++)
#define REPS(i, x) for (int i = 1; i <= x; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define REPZ(i, x) for (int i = 0; i <= x; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORA(i, n) for (auto &&i : n)
#define FORS(i, m, n) for (int i = m; i <= n; i++)
using namespace std;
#define PRINTARR(x, y)                                     \
  cerr << #x << "=\n";                                     \
  for (auto itr = x; itr != y; itr++) cerr << *itr << " "; \
  cerr << endl;
#define PRINTARR2(x, i0, i1)                                       \
  cerr << #x << "=\n";                                             \
  for (int ii0 = 0; ii0 < i0; ii0++) {                             \
    for (int ii1 = 0; ii1 < i1; ii1++) cerr << x[ii0][ii1] << " "; \
    cerr << endl;                                                  \
  }
#define DUMPOUT cerr
// vector
template <typename T> istream &operator>>(istream &is, vector<T> &vec) {
  for (T &x : vec) is >> x;
  return is;
}
// pair
template <typename T, typename U> ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
// vector
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "{";
  REP(i, (int)vec.size())
  os << vec[i] << (i + 1 == (int)vec.size() ? "" : ", ");
  os << "}";
  return os;
}
// map
template <typename T, typename U> ostream &operator<<(ostream &os, map<T, U> &map_var) {
  os << "{";
  FORA(itr, map_var) {
    os << *itr;
    itr++;
    if (itr != map_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}
// set
template <typename T> ostream &operator<<(ostream &os, set<T> &set_var) {
  os << "{";
  FORA(itr, set_var) {
    os << *itr;
    itr++;
    if (itr != set_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}
void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head &&head, Tail &&... tail) {
  DUMPOUT << head;
  if (sizeof...(Tail) > 0) DUMPOUT << ", ";
  dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define DUMP(...)                                                             \
  DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                             \
          << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]" << endl \
          << "    ",                                                          \
      dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define DUMP(...)
#endif

#define ALL(v) v.begin(), v.end()
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define epb emplace_back
#define int long long
#define pint pair<int, int>
#define ld long double
using namespace std;
template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> using vec = std::vector<T>;
template <class T> void print(const T &x) { cout << x << "\n"; }
const int MOD = 1000000007, INF0 = 1061109567, INF = INF0 * INF0;
const double EPS = 1e-10, PI = acos(-1.0);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
#define MAXN 2020
//----//
int N, K;
vec<int> G[MAXN];
int d[MAXN][MAXN];
struct edge {
  int from, to;
};
vec<edge> E;

void dfs(int id, int parent, int depth, int index) {
  d[index][id] = depth;
  FORA(e, G[id]) {
    if (e != parent) {
      dfs(e, id, depth + 1, index);
    }
  }
}
signed main() {
  cin.tie(0), ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  cin >> N >> K;
  REP(i, N - 1) {
    int p, q;
    cin >> p >> q;
    p--, q--;
    G[p].pb(q);
    G[q].pb(p);
    //    DUMP(p, q);
    edge e0 = {p, q};
    //  edge e1={q,p};
    E.pb(e0);
  }
  // PRINTARR(G[0].begin(), G[0].end());
  //点で確かめる
  int ans = INF;
  REP(i, N) dfs(i, -1, 0, i);

  if (K % 2 == 0) {
    REP(i, N) {
      int cnt = 0;
      REP(j, N) {
        if (d[i][j] > K / 2) cnt++;
      }
      chmin(ans, cnt);
    }
    print(ans);
  } else {
    REP(i, N - 1) {
      int cnt = 0;
      int a0 = E[i].from;
      int a1 = E[i].to;
      REP(j, N) {
        if (d[a0][j] > K / 2 && d[a1][j] > K / 2) cnt++;
      }
      chmin(ans, cnt);
    }
    print(ans);
  }
}