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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.AbstractMap.SimpleEntry;

class Main {
    static final int MOD = 1000000007;
    static final int INF = 1 << 30;

    Main() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int K = sc.nextInt();
        UnionFindTree uf = new UnionFindTree(N);
        List<SimpleEntry<Integer, Integer>> friendlist = new ArrayList<SimpleEntry<Integer, Integer>>(M);
        // List<SimpleEntry<Integer, Integer>> blocklist = new
        // ArrayList<SimpleEntry<Integer, Integer>>(K);
        int[] root = new int[N];
        int[] psize = new int[N];
        int[] infriend = new int[N];
        int[] inblock = new int[N];
        for (int i = 0; i < M; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            // g.doubleConnect(a, b);
            uf.union(a, b);
            friendlist.add(new SimpleEntry<Integer, Integer>(Math.min(a, b), Math.max(a, b)));
        }

        for (int i = 0; i < N; i++) {
            root[i] = uf.find(i);
            psize[root[i]]++;
        }

        for (int i = 0; i < K; i++) {
            int c = sc.nextInt() - 1;
            int d = sc.nextInt() - 1;
            if (root[c] == root[d]) {
                inblock[c]++;
                inblock[d]++;
            }
        }

        for (SimpleEntry<Integer, Integer> entry : friendlist) {
            if (root[entry.getKey()] == root[entry.getValue()]) {
                infriend[entry.getKey()]++;
                infriend[entry.getValue()]++;
            }
        }

        PrintWriter out = new PrintWriter(System.out);
        for (int i = 0; i < N; i++) {
            int val = psize[root[i]] - infriend[i] - inblock[i] - 1;
            // System.out.print(val + " ");
            // ans += val + " ";
            out.print(val);
            out.print(" ");
        }
        // System.out.println(ans);
        out.flush();
    }

    public static void main(String[] args) {
        new Main();
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public ArrayList<Integer> nextIntList(int n) {
        ArrayList<Integer> a = new ArrayList<Integer>(n);
        for (int i = 0; i < n; i++)
            a.add(nextInt());
        return a;
    }

    public ArrayList<Long> nextLongList(int n) {
        ArrayList<Long> a = new ArrayList<Long>(n);
        for (int i = 0; i < n; i++)
            a.add(nextLong());
        return a;
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}

class UnionFindTree {
    int[] parent;
    int[] rank;

    public UnionFindTree(int size) {
        this.parent = new int[size + 1];
        this.rank = new int[size + 1];
        for (int i = 0; i <= size; i++) {
            parent[i] = i;
            rank[i] = 0;
        }
    }

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

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

    public int find(int i) {
        if (i != parent[i]) {
            parent[i] = find(parent[i]);
        }
        return parent[i];
    }

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