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 {

    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    int k = sc.nextInt();

    public static void main(String[] args) {

        new Main().run();

    }

    void run() {
        int[] friendNum = new int[n];
        int[] blockNum = new int[n];

        UnionFind u = new UnionFind(n);

        for (int i=0; i<m; i++) {
            int a = sc.nextInt()-1;
            int b = sc.nextInt()-1;
            u.unite(a, b);
            friendNum[a]++;
            friendNum[b]++;
        }

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

            if (u.same(a, b)) {
                blockNum[a]++;
                blockNum[b]++;
            }
        }

        sc.close();

        StringBuilder sb = new StringBuilder();

        for (int i=0; i<n; i++) {
            int ans = u.size(i)-friendNum[i]-blockNum[i]-1;
            sb.append(ans+" ");
        }

        System.out.println(sb.toString());
    }

    class UnionFind {

        int[] parents, sizes;

        UnionFind(int n) {
            parents = new int[n];
            sizes = new int[n];
            for (int i=0; i<n; i++) {
                parents[i] = i;
                sizes[i] = 1;
            }
        }

        public int find(int x) {
            if (x==parents[x]) return x;
            return parents[x] = find(parents[x]);
        }

        public void unite(int x, int y) {
            x = find(x);
            y = find(y);
            if (x==y) return;
            if (sizes[x] < sizes[y]) {
                int tmp = x;
                x = y;
                y = tmp;
            }
            parents[y] = x;
            sizes[x] += sizes[y];
        }

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

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

    }
}