You are given an undirected tree, consisting of n vertices.

The vertex is called a leaf if it has exactly one vertex adjacent to it.

The distance between some pair of vertices is the number of edges in the shortest path between them.

Let's call some set of leaves beautiful if the maximum distance between any pair of leaves in it is less or equal to k.

You want to split all leaves into non-intersecting beautiful sets. What is the minimal number of sets in such a split?

Input

The first line contains two integers n and k (3 ≤ n ≤ 10^6, 1 ≤ k ≤ 10^6) — the number of vertices in the tree and the maximum distance between any pair of leaves in each beautiful set.

Each of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n) — the description of the i-th edge. 

It is guaranteed that the given edges form a tree.

Output

Print a single integer — the minimal number of beautiful sets the split can have. 

Examples

Input

9 3
1 2
1 3
2 4
2 5
3 6
6 7
6 8
3 9


Output

2


Input

5 3
1 2
2 3
3 4
4 5


Output

2


Input

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


Output

5

Note

Here is the graph for the first example:

<image>
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void in(T &x) {
  x = 0;
  char c = getchar();
  bool f = 0;
  while (!isdigit(c)) f |= (c == '-'), c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ '0'), c = getchar();
  f ? x = -x : 0;
}
template <class T>
inline void out(T x, const char c = '\n') {
  static short st[30];
  short m = 0;
  if (x < 0) putchar('-'), x = -x;
  do st[++m] = x % 10, x /= 10;
  while (x);
  while (m) putchar(st[m--] | '0');
  putchar(c);
}
template <class T>
inline void err(const T &x, const char c = '\n') {
  cerr << x << c;
}
template <class T, class... Args>
inline void in(T &x, Args &...args) {
  in(x);
  in(args...);
}
template <class T, class... Args>
inline void out(const T &x, const Args &...args) {
  out(x, ' ');
  out(args...);
}
template <class T, class... Args>
inline void err(const T &x, const Args &...args) {
  err(x, ' ');
  err(args...);
}
template <class T>
inline void prt(T a[], int n) {
  for (register int i = 0; i < n; ++i) out(a[i], i == n - 1 ? '\n' : ' ');
}
template <class T>
inline void clr(T a[], int n) {
  memset(a, 0, sizeof(T) * n);
}
template <class T>
inline void clr(T *a, T *b) {
  memset(a, 0, sizeof(T) * (b - a));
}
template <class T>
inline bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
inline bool ckmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
namespace MOD_CALC {
const int md = 66662333, inv2 = (md + 1) / 2;
inline int add(const int a, const int b) {
  return a + b >= md ? a + b - md : a + b;
}
inline int sub(const int a, const int b) {
  return a - b < 0 ? a - b + md : a - b;
}
inline int mul(const int a, const int b) { return (long long)a * b % md; }
inline void inc(int &a, const int b) { (a += b) >= md ? a -= md : 0; }
inline void dec(int &a, const int b) { (a -= b) < 0 ? a += md : 0; }
inline int qpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) r = mul(r, a);
  return r;
}
inline int qpow(int a, int b, const int p) {
  int r = 1;
  for (; b; b >>= 1, a = (long long)a * a % p)
    if (b & 1) r = (long long)r * a % p;
  return r;
}
inline int mdinv(const int a) { return qpow(a, md - 2); }
template <class... Args>
inline int add(const int a, const int b, const Args &...args) {
  return add(add(a, b), args...);
}
template <class... Args>
inline int mul(const int a, const int b, const Args &...args) {
  return mul(mul(a, b), args...);
}
}  // namespace MOD_CALC
using namespace MOD_CALC;
namespace i207M {
int n, k;
vector<int> E[1000005];
int ans;
int dfs(int x, int _fa) {
  vector<int> a;
  for (const auto &v : E[x]) {
    if (v == _fa) continue;
    a.push_back(dfs(v, x) + 1);
  }
  if (a.empty()) return 0;
  sort(a.begin(), a.end());
  int p = ((int)a.size()) - 1;
  while (p >= 1 && a[p] + a[p - 1] > k) ++ans, --p;
  return a[p];
}
signed main() {
  in(n, k);
  for (register int i = 1, a, b; i < n; ++i) {
    in(a, b);
    E[a].push_back(b), E[b].push_back(a);
  }
  int rt = 1;
  while (((int)E[rt].size()) == 1) ++rt;
  dfs(rt, 0);
  out(ans + 1);
  return 0;
}
}  // namespace i207M
signed main() {
  i207M::main();
  return 0;
}
