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 v5.12) (judge verdict: Time Limit Exceeded)

var input = require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");
var obj = {
  "list" : input,
  "index" : 0,
  "max" : input.length,
  "next" : function(){
    if(!this.hasNext()){throw "NoSuchElementException:次に要素は無いよ";}
    var returnObj = this.list[this.index];
    this.index++;
    return returnObj;
  },
  "hasNext" : function(){return (this.index < this.max);}
}
Main();
function next(){return obj.next();}
function hasNext(){return obj.hasNext();}
function myout(t){console.log(t);}//standard output
function myerr(t){console.error(t);}//standard error
//[no]要素の扱い。数値型
//不明値、異常時:引数そのまま返す　1:数値へ変換
//2:半角SPで分割 4:半角SPで分割し、数値配列へ
//6:1文字で分割  7:1文字で分割し、数値配列へ
//8:半角SPで結合 9:改行で結合 0:文字なしで結合
function myconv(i,no){try{switch(no){case 1:return parseInt(i);case 2:return i.split(" ");case 4:return i.split(" ").map((a)=>Number(a));case 6:return i.split("");case 7:return i.split("").map((a)=>Number(a));case 8:return i.join(" ");case 9:return i.join("\n");case 0:return i.join("");default:return i;}}catch(e){return i;}}
function Main(){
  var one = myconv(next(),4);
  var N = one[0];
  var M = one[1];
  var K = one[2];
  var list = new Array(N);
  var i;
  for(i = 0; i < N; i++){
    list[i] = {
    	"friend" : [],
        "group" : null
    };
  }
  for(i = 0; i < M; i++){
    var tmp = myconv(next(),4);
    list[tmp[0] - 1]["friend"].push(tmp[1] - 1);
    list[tmp[1] - 1]["friend"].push(tmp[0] - 1);
  }
  var group = [];
  for(i = 0; i < N; i++){
    bfs(i);
  }
  //myerr(list);
  function bfs(index){
    if(list[index]["group"] != null){
       return;
    }
    group.push(new Set());
  	var queue = [list[index]];
    while(queue.length > 0){
        var tmp = queue.shift();
      	if(tmp["group"] != null){
           continue;
        }
      	tmp["group"] = group.length - 1;
        for(var j = 0; j < tmp["friend"].length; j++){
          queue.push(list[tmp["friend"][j]]);
        }
    }
  }
  for(i = 0; i < N; i++){
    group[list[i]["group"]].add(i);
  }
  //myerr(group);
  var output = new Array(N);
  for(i = 0; i < N; i++){
    output[i] = group[list[i]["group"]].size - 1 - list[i]["friend"].length;
  }
  for(i = 0; i < K; i++){
    var ng = myconv(next(),4);
    ng[0]--;
    ng[1]--;
    if(list[ng[0]]["group"] != list[ng[1]]["group"]){
       continue;
    }
    output[ng[0]]--;
    output[ng[1]]--;
  }
  myout(myconv(output,8));
}