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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

//Don't have to see. start------------------------------------------
var read = require('readline').createInterface({
  input: process.stdin, output: process.stdout
});
var obj; var inLine = [];
read.on('line', function(input){inLine.push(input);});
read.on('close', function(){
  obj = init(inLine);
  console.error("\n");
  var start = Date.now();
  Main();
  var end = Date.now() - start;
  myerr("time : " + (end) + "ms");
  myerr("memory : " + Math.round(process.memoryUsage().heapTotal / 1024) + "KB");
});
function makeClone(obj){return JSON.parse(JSON.stringify(obj));}
function nextInt(){return myconv(next(),1);} function nextStrArray(){return myconv(next(),2);}
function nextIntArray(){return myconv(next(),4);} function nextCharArray(){return myconv(next(),6);}
function next(){return obj.next();} function hasNext(){return obj.hasNext();}
function init(input){  
  var returnObj = {
    list : input, index : 0, max : input.length,
    hasNext : function(){return (this.index < this.max);},
    next : function(){if(!this.hasNext()){throw "ArrayIndexOutOfBoundsException ‚There is no more input";}else{var returnInput = this.list[this.index++];return returnInput;}}
  };
  return returnObj;
}
function myout(s){console.log(s);}
function myerr(s){console.error("debug:" + require("util").inspect(s,false,null));}
//param "no" is
//unknown or outlier : return i. 1: parseInt.
//2: split space. 4: split space and parseInt.
//6: split 1 character. 7: split 1 character and parseInt.
//8: join space. 9: join nextline. 0: join no character.
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map(Number);case 6:return i.split("");case 7:return i.split("").map(Number);case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}

//Don't have to see. end------------------------------------------
function Main(){
  var one = nextIntArray();
  var N = one[0];
  var K = one[1];
  var list = new Array(N);
  for(var i = 0; i < N; i++){
    list[i] = {
    	access : false,
      	child : [],
      	depth : 0
    };
  }
  for(var i = 0; i < N - 1; i++){
    var tmp = nextIntArray();
    list[tmp[0] - 1].child.push(tmp[1] - 1);
    list[tmp[1] - 1].child.push(tmp[0] - 1);
  }
  var output = 5000;
  for(var i = 0; i < N; i++){
    if(list[i].child.length == 1){
      output = Math.min(output, bfs(i));
    }else{
      continue;
    }
  }
  myout(output);
  function bfs(s){
    for(var i = 0; i < N; i++){
      list[i].access = false;
      list[i].depth = 0;
    }
    var count = 0;
    var queue = [s];
    list[s].access = true;
    while(queue.length > 0){
      var now = queue.shift();
      var child = list[now].child;
      if(list[now].depth > K){
        count++;
      }
      for(var i = 0; i < child.length; i++){
        if(!list[child[i]].access){
          list[child[i]].access = true;
          list[child[i]].depth = list[now].depth + 1;
          queue.push(child[i]);
        }
      }
    }
    return count;
  }
}