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: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    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[] A = new int[M];
        int[] B = new int[M];
        for (int i=0; i<M; i++) {
            A[i] = sc.nextInt();
            B[i] = sc.nextInt();
        }
        int[] C = new int[K];
        int[] D = new int[K];
        for (int i=0; i<K; i++) {
            C[i] = sc.nextInt();
            D[i] = sc.nextInt();
        }

        int[] ans = solve(N, M, K, A, B, C, D);
        for (int i=0; i<N; i++) {
            if (i != 0) {
                System.out.print(" ");
            }
            System.out.print(ans[i]);
        }
        System.out.println();
    }

    private static int[] solve(int N, int M, int K, int[] A, int[] B, int[] C, int[] D) {
        int[] ans = new int[N];

        UnionFind uf = new UnionFind(N);
        int[] blocked = new int[N+1];
        int[] friend = new int[N+1];

        for (int i=0; i<M; i++) {
            uf.unite(A[i], B[i]);
            friend[A[i]]++;
            friend[B[i]]++;
        }
        for (int i=0; i<K; i++) {
            if (uf.isSame(C[i], D[i])) {
                blocked[C[i]]++;
                blocked[D[i]]++;
            }
        }

        for (int i=0; i<N; i++) {
            ans[i] = uf.size(i+1) - friend[i+1] - blocked[i+1] - 1;
            System.out.println("i=" + i + ": " + uf.size(i+1) + " - " + friend[i+1] + " - " + blocked[i+1] + " - 1 = " + ans[i]);
        }

        return ans;
    }


    public static class UnionFind {
        /**
         * 各要素のrootを示す．対象要素がrootの場合は木の要素数を負の値で表現する
         */
        int[] parent;

        public UnionFind(int N) {
            parent = new int[N+1];
            Arrays.fill(parent, -1);
        }

        /**
         * xとyの木を併合する
         * @param x
         * @param y
         */
        public void unite(int x, int y) {
            int rootx = find(x);
            int rooty = find(y);

            // XとYが同一木なら何もしない
            if (rootx == rooty) return;

            // 小さい木を大きい木に併合したいので
            // Xの方が小さい場合はXとYを入れ替えてYをXに併合する
            if (size(x) < size(y)) {
                unite(y, x);
                return;
            }

            parent[rootx] += parent[rooty];
            parent[rooty] = x;
        }

        /**
         * UnionFind木のルートを求める
         * @param x
         * @return
         */
        public int find(int x){
            if (parent[x] < 0) {
                return x;
            }

            return parent[x] = find(parent[x]);
        }

        /**
         * 2つの要素が同一グループに属するか
         * @param x
         * @param y
         * @return
         */
        public boolean isSame(int x, int y) {
            return find(x) == find(y);
        }

        public int size(int x) {
            return -parent[find(x)];
        }
    }


}