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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main{
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int m = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        
        UnionFind uf = new UnionFind(n);
        List<Integer>[] graph = new ArrayList[n];
        for(int i=0; i<n; i++){
            graph[i] = new ArrayList<>();
        }
        Map<Long,Boolean> block = new HashMap<>();
        
        for(int i=0; i<m; i++){
            int a = Integer.parseInt(sc.next())-1;
            int b = Integer.parseInt(sc.next())-1;
            graph[a].add(b);
            graph[b].add(a);
            uf.unite(a,b);
        }
        
        int[] ans = new int[n];
        for(int i=0; i<n; i++){
            ans[i] = uf.size(i)-graph[i].size();
        }
        
        for(int i=0; i<k; i++){
            int c = Integer.parseInt(sc.next())-1;
            int d = Integer.parseInt(sc.next())-1;
            if(uf.same(c,d)){
                ans[c]--;
                ans[d]--;
            }
        }
        
        for(int i=0; i<n; i++){
            System.out.println(ans[i]-1);
        }
        
    }
}

class UnionFind{
    int n;
    int[] par;
    int[] size;
    
    public UnionFind(int n){
        this.n = n;
        par = new int[n];
        for(int i=0; i<n; i++){
            par[i] = i;
        }
        size = new int[n];
        Arrays.fill(size, 1);
    }
    
    public int find(int x){
        if(par[x] == x){
            return x;
        }else{
            par[x] = find(par[x]);
            return par[x];
        }
    }
    
    public boolean unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x==y) return false;
        
        if(size[x]<size[y]){
            int tmp = x;
            x = y;
            y = tmp;
        }
        size[x] += size[y];
        par[y] = x;
        return true;
    }
    
    public boolean same(int x, int y){
        return find(x)==find(y);
    }
    
    public int size(int x){
        return size[find(x)];
    }
    
    public int groupCount(){
        Set<Integer> s = new HashSet<>();
        for(int i=0; i<n; i++){
            s.add(this.find(i));
        }
        return s.size();
    }
}