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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ll = long long int;
using vll = vector<ll>; using vvll = vector<vll>; using vvvll = vector<vvll>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;

const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
#define Loop(i, n) for(int i = 0; i < (int)(n); i++)
#define Loop1(i, n) for(int i = 1; i <= (int)(n); i++)
#define Loopr(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)(n); i >= 1; i--)
#define Foreach(buf, container) for(auto buf : container)
#define Loopdiag(i, j, h, w, sum) for(int i = ((sum) >= (h) ? (h) - 1 : (sum)), j = (sum) - i; i >= 0 && j < (w); i--, j++)
#define Loopdiagr(i, j, h, w, sum) for(int j = ((sum) >= (w) ? (w) - 1 : (sum)), i = (sum) - j; j >= 0 && i < (h); j--, i++)
#define Loopdiagsym(i, j, h, w, gap) for (int i = ((gap) >= 0 ? (gap) : 0), j = i - (gap); i < (h) && j < (w); i++, j++)
#define Loopdiagsymr(i, j, h, w, gap) for (int i = ((gap) > (h) - (w) - 1 ? (h) - 1 : (w) - 1 + (gap)), j = i - (gap); i >= 0 && j >= 0; i--, j--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define printv(vector) Loop(ex_i, vector.size()) { cout << vector[ex_i] << " "; } cout << endl;
#define printmx(matrix) Loop(ex_i, matrix.size()) { Loop(ex_j, matrix[ex_i].size()) { cout << matrix[ex_i][ex_j] << " "; } cout << endl; }
#define quickio() ios::sync_with_stdio(false); cin.tie(0);
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
#define Comp(type_t) bool operator<(const type_t &another) const
#define fst first
#define snd second
bool nearlyeq(double x, double y) { return abs(x - y) < 1e-9; }
bool inrange(int x, int t) { return x >= 0 && x < t; }
bool inrange(vi xs, int t) { Foreach(x, xs) if (!(x >= 0 && x < t)) return false; return true; }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (m * m <= x ? 0 : -1); }
ll ceilsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }
ll lcm(ll m, ll n) { return m * n / gcd(m, n); }

/*******************************************************/

int n;
int k;
bool judge;
vvi lst;

P solve(int a, int to, int from, int d) {
	P ret = { 0, -1 };
	vi ds;
	if (d > k) ret.fst++;
	else ret.snd = d;
	if (to != -1) {
		Foreach(b, lst[a]) {
			if (b != to) {
				P buf = solve(b, -1, a, n);
				ret.fst += buf.fst;
			}
		}
		P buf = solve(to, -1, a, d + 1);
		ret.fst += buf.fst;
		if (buf.snd != -1) {
			ret.snd = max(ret.snd, buf.snd);
			ds.push_back(buf.snd);
		}
	}
	else {
		Foreach(b, lst[a]) {
			if (b != from) {
				P buf = solve(b, -1, a, d + 1);
				ret.fst += buf.fst;
				if (buf.snd != -1) {
					ret.snd = max(ret.snd, buf.snd);
					ds.push_back(buf.snd);
				}
			}
		}
	}
	sort(ds.begin(), ds.end(), greater<int>());
	if (ds.size() >= 2) {
		if (ds[0] - d + ds[1] - d > k) judge = false;
	}
	return ret;
}

int main() {
	quickio();
	cin >> n >> k;
	lst = vvi(n);
	Loop(i, n - 1) {
		int s, t; cin >> s >> t; s--; t--;
		lst[s].push_back(t);
		lst[t].push_back(s);
	}
	int ans = INT_MAX;
	Loop(i, n) {
		Foreach(b, lst[i]) {
			judge = true;
			int buf = solve(i, b, -1, 0).fst;
			if (judge) ans = min(ans, buf);
		}
	}
	cout << ans << endl;
}