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)

// inputに入力データ全体が入る
function Main(input) {
    var N = input[0].split(" ")[0]-0;
    var M = input[0].split(" ")[1]-0;
    var K = input[0].split(" ")[2]-0;
    
    var uf = new UnionFind(N);
    var map = {};
    for(var i=0;i<M;i++){
        var a = input[i+1].split(" ")[0]-1;
        var b = input[i+1].split(" ")[1]-1;
        uf.link(a,b);
        if(!map[a])map[a]={};
        if(!map[b])map[b]={};
        map[a][b] = false;
        map[b][a] = false;
    }
    for(var i=0;i<K;i++){
        var a = input[i+M+1].split(" ")[0]-1;
        var b = input[i+M+1].split(" ")[1]-1;
        if(!map[a]) map[a] = {};
        if(!map[b]) map[b] = {};
        map[a][b] = false;
        map[b][a] = false;
    }
    var fmap = {};
    var cmap = {};
    for(var i=0;i<N;i++){
        var f = uf.find(i);
        if(!fmap[f]) fmap[f] = {};
        fmap[f][i] = true;
        cmap[i] = fmap[f];
        if(!map[i]) map[i] = {};
        map[i][i]=false;
    }
    
    var ans = [];
    for(var i=0;i<N;i++){
        ans[i] = 0;
        if(cmap[i]) Object.keys(cmap[i]).forEach(key=>{
            if(map[i][key]!==false) ans[i]++;
        });
    }
    console.log(ans.join(" "));
}

function UnionFind(count) {
  this.roots = new Array(count);
  this.ranks = new Array(count);
  
  for(var i=0; i<count; ++i) {
    this.roots[i] = i;
    this.ranks[i] = 0;
  }
}

var proto = UnionFind.prototype

Object.defineProperty(proto, "length", {
  "get": function() {
    return this.roots.length
  }
})

proto.makeSet = function() {
  var n = this.roots.length;
  this.roots.push(n);
  this.ranks.push(0);
  return n;
}

proto.find = function(x) {
  var x0 = x
  var roots = this.roots;
  while(roots[x] !== x) {
    x = roots[x]
  }
  while(roots[x0] !== x) {
    var y = roots[x0]
    roots[x0] = x
    x0 = y
  }
  return x;
}

proto.link = function(x, y) {
  var xr = this.find(x)
    , yr = this.find(y);
  if(xr === yr) {
    return;
  }
  var ranks = this.ranks
    , roots = this.roots
    , xd    = ranks[xr]
    , yd    = ranks[yr];
  if(xd < yd) {
    roots[xr] = yr;
  } else if(yd < xd) {
    roots[yr] = xr;
  } else {
    roots[yr] = xr;
    ++ranks[xr];
  }
}


Main(require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n"));