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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt(); // ninzu
        int M = sc.nextInt(); // firend
        int K = sc.nextInt(); // block

        int[] kouho = new int[N];
        UnionFindTree uft = new UnionFindTree(N);

        for (int i = 0; i < M; i++) {
            int A = sc.nextInt() - 1;
            int B = sc.nextInt() - 1;
            kouho[A]--;
            kouho[B]--;
            uft.union(A, B);
        }

        for (int i = 0; i < K; i++) {
            int C = sc.nextInt() - 1;
            int D = sc.nextInt() - 1;
            if (uft.same(C, D)) {
                kouho[C]--;
                kouho[D]--;
            }
        }

        for (int i = 0; i < N; i++) {
            if (i != 0) {
                System.out.print(" ");
            }

            for (int j = 0; j < N; j++) {
                if (i == j) {
                    continue;
                }

                if (uft.same(i, j)) {
                    kouho[i]++;
                }
            }
            System.out.print(kouho[i]);
        }
    }

}

//copied by :https://gist.github.com/keitin/e5ba4ba4ccd6cd3afb3485c5ea85846c#file-union-find-java
class UnionFindTree {

    int[] parent; //インデックスにとノードを対応させ、そのルートノードのインデックスを格納 
    int[] rank; //parentと同様に、木の高さを格納

    public UnionFindTree(int size) {
        this.parent = new int[size];
        this.rank = new int[size];

        for (int i = 0; i < size; i++) {
            makeSet(i);
        }
    }

    public void makeSet(int i) {
        parent[i] = i;
        rank[i] = 0; //集合の高さ
    }

    public void union(int x, int y) {
        int xRoot = find(x);
        int yRoot = find(y);

        //xが属する木の方が大きい場合
        if (rank[xRoot] > rank[yRoot]) {
            parent[yRoot] = xRoot; //yの親をxに更新 

        } else if (rank[xRoot] < rank[yRoot]) {
            parent[xRoot] = yRoot;

        } else if (xRoot != yRoot) {
            parent[yRoot] = xRoot;
            rank[xRoot]++; //同じ高さの木がルートの子として着くから大きさ++;
        }
    }

    //引数の属する木のルートのidを返す
    public int find(int i) {
        if (i != parent[i]) {
            parent[i] = find(parent[i]);
        }
        return parent[i];
    }

    //同じ木に属しているか
    public boolean same(int x, int y) {
        return find(x) == find(y);
    }
}