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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }

    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int M = scan.nextInt();
        int K = scan.nextInt();
        HashMap<Integer, List<Integer>> real = new HashMap<>();
        UnionFind good = new UnionFind(N);
        for (int i = 0; i < M; i++) {
            int a = scan.nextInt()-1;
            int b = scan.nextInt()-1;
            if (!real.containsKey(a)) {
                real.put(a, new ArrayList<>());
            }
            if (!real.containsKey(b)) {
                real.put(b, new ArrayList<>());
            }
            real.get(a).add(b);
            real.get(b).add(a);
            good.unit(a, b);
        }
        HashMap<Integer, List<Integer>> bad = new HashMap<>();
        for (int i = 0; i < K; i++) {
            int a = scan.nextInt()-1;
            int b = scan.nextInt()-1;
            if (!good.same(a, b)) {
                continue;
            }
            if (!bad.containsKey(a)) {
                bad.put(a, new ArrayList<>());
            }
            if (!bad.containsKey(b)) {
                bad.put(b, new ArrayList<>());
            }
            bad.get(a).add(b);
            bad.get(b).add(a);
        }
        int[] ans = new int[N];
        for (int i = 0; i < N; i++) {
            int candidates = good.size(i) - 1 - real.getOrDefault(i, new ArrayList<>()).size() - bad.getOrDefault(i, new ArrayList<>()).size();
            ans[i] = candidates;
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < N; i++) {
            sb.append(ans[i]);
            sb.append(" ");
        }
        System.out.println(sb.substring(0, sb.length()-1));
    }
    class UnionFind {
        private int[] parent;
        private int[] size;
        UnionFind(int N) {
            int[] parent = new int[N];
            int[] size = new int[N];
            Arrays.fill(size, 1);
            for (int i = 0; i < N; i++) {
                parent[i] = i;
            }
            this.parent = parent;
            this.size = size;
        }
        int find(int x) {
            if (this.parent[x] == x) {
                return x;
            }
            return this.parent[x] = find(this.parent[x]);
        }
        boolean same(int x, int y) {
            return find(x) == find(y);
        }
        void unit(int x, int y) {
            int parent_x = find(x);
            int parent_y = find(y);
            if (parent_x == parent_y) {
                return;
            }
            if (this.size[parent_x] < this.size[parent_y]) {
                this.parent[parent_x] = parent_y;
                this.size[parent_y] += this.size[parent_x];
                this.size[parent_x] = 0;
                return;
            }
            this.parent[parent_y] = parent_x;
            this.size[parent_x] += this.size[parent_y];
            this.size[parent_y] = 0;
        }
        int size(int x) {
            int parent_x = find(x);
            return this.size[parent_x];
        }
    }
}