Problem:
Score : 400 points

Problem Statement

An SNS has N users - User 1, User 2, \cdots, User N.

Between these N users, there are some relationships - M friendships and K blockships.

For each i = 1, 2, \cdots, M, there is a bidirectional friendship between User A_i and User B_i.

For each i = 1, 2, \cdots, K, there is a bidirectional blockship between User C_i and User D_i.

We define User a to be a friend candidate for User b when all of the following four conditions are satisfied:

a \neq b.

There is not a friendship between User a and User b.

There is not a blockship between User a and User b.

There exists a sequence c_0, c_1, c_2, \cdots, c_L consisting of integers between 1 and N (inclusive) such that c_0 = a, c_L = b, and there is a friendship between User c_i and c_{i+1} for each i = 0, 1, \cdots, L - 1.

For each user i = 1, 2, ... N, how many friend candidates does it have?

Constraints

All values in input are integers.

2 ≤ N ≤ 10^5

0 \leq M \leq 10^5

0 \leq K \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

1 \leq C_i, D_i \leq N

C_i \neq D_i

(A_i, B_i) \neq (A_j, B_j) (i \neq j)

(A_i, B_i) \neq (B_j, A_j)

(C_i, D_i) \neq (C_j, D_j) (i \neq j)

(C_i, D_i) \neq (D_j, C_j)

(A_i, B_i) \neq (C_j, D_j)

(A_i, B_i) \neq (D_j, C_j)

Input

Input is given from Standard Input in the following format:

N M K
A_1 B_1
\vdots
A_M B_M
C_1 D_1
\vdots
C_K D_K

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User 2 and 3, and between 3 and 4. Also, there is no friendship or blockship between User 2 and 4. Thus, User 4 is a friend candidate for User 2.

However, neither User 1 or 3 is a friend candidate for User 2, so User 2 has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

//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↑入力 ↓出力');
  Main();
});
function makeClone(obj){return (obj instanceof Set) ? new Set(Array.from(obj)) : 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){  
  return {
    list : input, index : 0, max : input.length,
    hasNext : function(){return (this.index < this.max);},
    next : function(){if(this.hasNext()){return this.list[this.index++];}else{throw 'ArrayIndexOutOfBoundsException ‚There is no more input';}}
  };
}
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 M = one[1];
  var K = one[2];
  var uf = unionFind(N);
  var map = {};
  for(var i = 0; i < N; i++){
    map[i] = {
      ok : new Set(),
      ng : new Set()
    };
  }
  for(var i = 0; i < M; i++){
    var tmp = nextIntArray();
    tmp[0]--;
    tmp[1]--;
    uf.doUnion(tmp[0], tmp[1]);
    map[tmp[0]].ok.add(tmp[1]);
    map[tmp[1]].ok.add(tmp[0]);
  }
  for(var i = 0; i < K; i++){
    var tmp = nextIntArray();
    tmp[0]--;
    tmp[1]--;
    map[tmp[0]].ng.add(tmp[1]);
    map[tmp[1]].ng.add(tmp[0]);
  }
  var output = new Array(N).fill(0);
  for(var i = 0; i < N; i++){
    output[i] = uf.getSize(i) - map[i].ok.size - 1;
    var ngs = Array.from(map[i].ng);
    for(var j = 0; j < ngs.length; j++){
      if(uf.isSame(i, ngs[j])){
        output[i]--;
      }
    }
  }
  myout(myconv(output, 8));
}

function unionFind(n){
  var uf = {
    //全てのインデックスは「-X(親、絶対値はグループの大きさ)」「自分が属する親(≒根)」のいずれかを持つ。
    //最初はみんなグループの大きさ1の親
    "list" : new Array(n).fill(-1),
    
    //同じ親を持つか
    "isSame" : function(mae, ato){
      return this.getRootIndex(mae) == this.getRootIndex(ato);
    },
    //自身の親インデックスをたどって根っこに着く
    //親にたどり終えたら子に帰っていくついでに、子たちに「共通の親を持っている」ことを知らせる
    "getRootIndex" : function(index){
      if(this.list[index] < 0){
        return index;
      }else{
        this.list[index] = this.getRootIndex(this.list[index]);
        return this.list[index];
      }
    },
    //異なる親同士のまとまりを一つにする（同じ親ならスルー）
    //小さいグループの親が大きいグループの親の直下に着く
    //グループの大きさも更新する
    "doUnion" : function(mae, ato){
      var maeRoot = this.getRootIndex(mae);
      var atoRoot = this.getRootIndex(ato);
      if(!this.isSame(maeRoot, atoRoot)){
        if(maeRoot >= atoRoot){
          this.list[maeRoot] += this.list[atoRoot];
          this.list[atoRoot] = maeRoot;
        }else{
          this.list[atoRoot] += this.list[maeRoot];
          this.list[maeRoot] = atoRoot;
        }
      }
    },
    //「-X(親、絶対値はグループの大きさ)」
    //なので、インデックスを指定→親を知る→親の持つグループの大きさがわかる。
    //ただしマイナス値なので、掛け算して返す。
    "getSize" : function(index){
      return -this.list[this.getRootIndex(index)];
    }
  }
  return uf;
}