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: Runtime Error)

import java.util.*;
public class Main{
	public static int[] par;
    public static int[] sizes;

	public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      par = new int[n];
      sizes = new int[n];
      unionFind uf = new unionFind(n);
      int m = sc.nextInt();
      int k = sc.nextInt();
      int a,b,c,d;
      
      int[][] no = new int[n][m];
      int[] cnt=new int[n];
      for(int i=0;i<m;i++){
        a =sc.nextInt()-1;
        b =sc.nextInt()-1;
        uf.unite(a,b);
        no[a][cnt[a]++] = b;
        no[b][cnt[b]++] = a;
      }
      for(int i=0;i<k;i++){
        c = sc.nextInt()-1;
        d = sc.nextInt()-1;
        if(!(uf.same(c,d))) continue;
        no[c][cnt[c]++] = d;
        no[d][cnt[d]++] = c;
      }
      for(int i=0;i<n;i++){
        int ans = uf.size(i)-1-cnt[i];
        System.out.print(ans+" ");
      }
    }
}
  
class unionFind{
    unionFind(int n){
		for(int i=0;i<n;i++) Main.par[i]=i;
		for(int i=0;i<n;i++) Main.sizes[i]=1;
	}
	
	int root(int x){
		if(Main.par[x]==x) return x;  //xが根ならばxを返す
		return Main.par[x] = root(Main.par[x]);  //経路圧縮
	}
	
	void unite(int x,int y){
		int rx = root(x);
		int ry = root(y);
		if(rx == ry) return ;
		if(Main.sizes[rx]<Main.sizes[ry]){
			int temp;
			temp=rx;
			rx=ry;
			ry=temp;	
		}
		Main.par[ry] = rx;  //根rxと根ryが別物だったならryをrxに付ける
		Main.sizes[rx]+=Main.sizes[ry];
	}
	
	boolean same(int x,int y){
		int rx = root(x);
	    int ry = root(y);
	    return rx==ry;
	}

	int size(int x){
		return Main.sizes[root(x)];
	}
}