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.IOException;
import java.io.InputStream;
import java.util.HashMap;
import java.util.HashSet;
import java.util.NoSuchElementException;
import static java.util.Arrays.fill;


public class Main {
    public static void main(String[] args) {
        FastScanner fsc = new FastScanner();
        int n = fsc.nextInt();
        int m = fsc.nextInt();
        int k = fsc.nextInt();
        HashSet<Integer>[] friend = new HashSet[n];
        HashSet<Integer>[] block = new HashSet[n];
        for (int i = 0; i < n; i++) {
            block[i] = new HashSet<>();
            friend[i] = new HashSet<>();
        }
        UnionFindTree t = new UnionFindTree(n);
        for (int i = 0; i < m; i++) {
            int a = fsc.nextInt() - 1;
            int b = fsc.nextInt() - 1;
            friend[a].add(b);
            friend[b].add(a);
            t.unite(a, b);
        }
        for (int i = 0; i < k; i++) {
            int c = fsc.nextInt() - 1;
            int d = fsc.nextInt() - 1;
            if (t.root(c) == t.root(d)) {
                block[c].add(d);
                block[d].add(c);
            }
        }
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            if (!map.containsKey(t.root(i))) {
                map.put(t.root(i), 1);
            } else {
                map.put(t.root(i), map.get(t.root(i)) + 1);
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            sb.append(map.get(t.root(i)) - block[i].size() - friend[i].size() - 1).append(' ');
        }
        System.out.println(sb);
    }
}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}


class UnionFindTree {
    private int[] parent;
    private int[] rank;
    private int[] size;
    private int groups;

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

    public int root(int x) {
        if (parent[x] == x) {
            return x;
        }
        return parent[x] = root(parent[x]);
    }

    public boolean isSame(int x, int y) {
        return root(x) == root(y);
    }

    public void unite(int x, int y) {
        int xr = root(x);
        int yr = root(y);
        if (xr == yr) {
            return;
        }
        if (rank[xr] < rank[yr]) {
            parent[xr] = yr;
            size[yr] += size[xr];
        } else {
            parent[yr] = xr;
            size[xr] += size[yr];
            if (rank[xr] == rank[yr]) {
                rank[xr]++;
            }
        }
        groups--;
    }

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

    public boolean isRoot(int x) {
        return parent[x] == x;
    }

    public int groups() {
        return groups;
    }
}