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 = sc.nextInt();
        int M = sc.nextInt();
        int K = sc.nextInt();
        int[] roots = new int[N];
        int[] cnts = new int[N];
        int[] jisu = new int[N];
        int[] ans = new int[N];
        for(int i=0;i<N;i++){
            roots[i]=i;
            cnts[i] = 0;
            jisu[i]=0;
        }

        for(int i=0;i<M;i++){
            int a = sc.nextInt()-1;
            int b = sc.nextInt()-1;
            jisu[a]++;
            jisu[b]++;
            my_union(roots, a, b);
        }
        for(int i=0;i<N;i++){
            int root = my_find(roots, i);
            cnts[root]++;
        }
        for(int i=0;i<N;i++){
            int root = my_find(roots, i);
            ans[i] = cnts[root]-1;
            ans[i] -= jisu[i];
        }
        for(int i=0;i<K;i++){
            int c = sc.nextInt()-1;
            int d = sc.nextInt()-1;
            if(my_find(roots, c) == my_find(roots, d)){
                ans[c]--;
                ans[d]--;
            }
        }
        for(int p:ans){
            System.out.print(p + " ");
        }
    }

    private static void printarr(int[] a){
        for(int p:a){
            System.out.print(p);
        }
        System.out.println();
    }

    private static int my_find(int[] roots, int i){
        List<Integer> ch = new ArrayList<>();
        while(roots[i] != i){
            ch.add(i);
            i = roots[i];
        }
        for(Integer c:ch){
            roots[c] = i;
        }
        return i;
    }

    private static void my_union(int[] roots, int a, int b){
        int ra = my_find(roots, a);
        int rb = my_find(roots, b);
        roots[ra] = rb;
        return;
    }

}