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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Scanner;
import java.util.LinkedList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        D solver = new D();
        solver.solve(1, in, out);
        out.close();
    }

    static class D {
        static int ans[];

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 入力
            int n = Integer.parseInt(in.next());
            int m = Integer.parseInt(in.next());
            int k = Integer.parseInt(in.next());

            UnionFind uf = new UnionFind(n);

            ans = new int[n];

            int like[][] = new int[m][2];
            for (int i = 0; i < m; i++) {
                like[i][0] = Integer.parseInt(in.next()) - 1;
                like[i][1] = Integer.parseInt(in.next()) - 1;
                uf.unite(like[i][0], like[i][1]);
            }
            int dislike[][] = new int[k][2];
            for (int i = 0; i < k; i++) {
                dislike[i][0] = Integer.parseInt(in.next()) - 1;
                dislike[i][1] = Integer.parseInt(in.next()) - 1;
            }
            for (int i = 0; i < n; i++) {
                ans[i] = (int) uf.getSize(i) - 1;
            }
            for (int i = 0; i < m; i++) {
                ans[like[i][0]]--;
                ans[like[i][1]]--;
            }
            for (int i = 0; i < k; i++) {
                if (uf.same(dislike[i][0], dislike[i][1])) {
                    ans[dislike[i][0]]--;
                    ans[dislike[i][1]]--;
                }
            }
            for (int i = 0; i < n; i++) {
                out.printf("%d ", ans[i]);
            }
        }

        public class UnionFind {
            private List<Integer> parent = new LinkedList<Integer>();
            private List<Integer> rank = new LinkedList<Integer>();
            private List<Integer> elmsize = new LinkedList<Integer>();

            public UnionFind() {
            }

            public UnionFind(int size) {
                for (int i = 0; i < size; i++) {
                    parent.add(i);
                    rank.add(0);
                    elmsize.add(1);
                }
            }

            public void unite(int x, int y) {
                final int xRoot = root(x);
                final int yRoot = root(y);

                if (rank.get(xRoot) > rank.get(yRoot)) {
                    parent.set(yRoot, xRoot);    //x の木に結合
                    elmsize.set(xRoot, elmsize.get(xRoot) + elmsize.get(yRoot));
                } else if (rank.get(xRoot) < rank.get(yRoot)) {
                    parent.set(xRoot, yRoot);    //y の木に結合
                    elmsize.set(yRoot, elmsize.get(yRoot) + elmsize.get(xRoot));
                } else if (xRoot != yRoot) {        //同じ高さで根が違うとき
                    parent.set(yRoot, xRoot);    //x の木に結合
                    rank.set(xRoot, rank.get(xRoot) + 1);    //x の木の高さを加算
                    elmsize.set(xRoot, elmsize.get(xRoot) + elmsize.get(yRoot));
                }
            }

            public int root(int x) {
                if (x != parent.get(x)) {    //根でないとき
                    parent.set(x, root(parent.get(x)));    //直接、根に繋ぐ(経路圧縮)
                }
                return parent.get(x);
            }

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

            public long getSize(int x) {
                return elmsize.get(root(x));
            }

        }

    }
}