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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            int M = sc.nextInt();
            int K = sc.nextInt();

            UnionFindTree uft = new UnionFindTree(N);

            List<Integer>[] friends = new List[N];
            for (int i = 0; i < N; i++) {
                friends[i] = new ArrayList<>();
            }

            for (int i = 0; i < M; i++) {
                int a = sc.nextInt() - 1;
                int b = sc.nextInt() - 1;

                friends[a].add(b);
                friends[b].add(a);

                uft.unite(a, b);
            }

            List<Integer>[] groups = new List[N];
            for (int i = 0; i < N; i++) {
                groups[i] = new ArrayList<>();
            }

            for (int i = 0; i < N; i++) {
                groups[uft.find(i)].add(i);
            }

            int[] C = new int[N];
            for (int i = 0; i < N; i++) {
                C[i] = groups[uft.find(i)].size() - 1 - friends[i].size();
            }

            for (int i = 0; i < K; i++) {
                int c = sc.nextInt() - 1;
                int d = sc.nextInt() - 1;

                if (uft.find(c) == uft.find(d)) {
                    C[c]--;
                    C[d]--;
                }
            }

            StringBuilder ans = new StringBuilder();
            for (int i = 0; i < N; i++) {
                ans.append(C[i]).append(" ");
            }

            System.out.println(ans.toString().trim());
        }
    }

    public static class UnionFindTree {

        private int[] parent;

        private int[] rank;

        public UnionFindTree(int N) {
            parent = new int[N];
            rank = new int[N];
            for (int i = 0; i < N; i++) {
                parent[i] = i;
                rank[i] = 0;
            }
        }

        public int find(int x) {
            if (parent[x] == x) {
                return x;
            } else {
                int p = find(parent[x]);
                parent[x] = p;
                return p;
            }
        }

        public void unite(int x, int y) {
            x = find(x);
            y = find(y);

            if (x == y) {
                return;
            }

            if (rank[x] < rank[y]) {
                parent[x] = y;
            } else {
                parent[y] = x;
                if (rank[x] == rank[y]) {
                    rank[x]++;
                }
            }
        }

        public boolean same(int x, int y) {
            return find(x) == find(y);
        }

    }

}