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.*;
import java.io.*;
import java.math.BigInteger;

public class Main {
	public static class pair implements Comparable<pair> {
		int x;
		int y;

		public pair(int par1, int par2) {
			x = par1;
			y = par2;
		}

		@Override
		public int compareTo(pair p) {
			return x - p.x;
		}

		@Override
		public String toString() {
			return +x + " " + y;
		}

	}

	static long mod = (long) (1e9 + 7);

	public static long facn(long n, long x) {
		long ans = 1;
		for (int i = 1; i <= x; i++) {
			ans *= n;
			ans %= mod;
			n--;
		}
		long k = 1;
		long t = x;
		for (int i = 1; i <= t; i++) {
			k *= x;
			k %= mod;
			x--;
		}
//		BigInteger b = BigInteger.valueOf((k)).modInverse(BigInteger.valueOf((mod)));
		long b = modPow(k, mod - 2, mod);
		return ((b % mod) * ans) % mod;

	}

	public static long modInverse(long a, long m) {
		a = a % m;
		for (int x = 1; x < m; x++)
			if ((a * x) % m == 1)
				return x;
		return 1;
	}

	static long modPow(long a, long e, long mod) // O(log e)
	{
		a %= mod;
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				res = (res * a) % mod;
			a = (a * a) % mod;
			e >>= 1;
		}
		return res % mod;
	}

	public static void bb(Serializable s) {
		System.out.println(55);
	}

	public static int dfs(int u) {
		vis[u] = true;
		rank[u] = rnk;
		int ans = 1;
		for (int v : adj[u])
			if (!vis[v])
				ans += dfs(v);

		return ans;
	}

	public static void dfsD(int u) {
		dist[u] = dst;
		for (int v : adj[u])
			if (dist[v] == -1)
				dfsD(v);
	}

	static ArrayList<Integer>[] adj, adj2;
	static int n, m, k, rnk, dst;
	static boolean[] vis;
	static int[] dist, rank;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		n = sc.nextInt();
		m = sc.nextInt();
		k = sc.nextInt();
		adj = new ArrayList[n];
		adj2 = new ArrayList[n];
		vis = new boolean[n];
		dist = new int[n];
		Arrays.fill(dist, -1);
		rank = new int[n];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList<Integer>();
		for (int i = 0; i < n; i++)
			adj2[i] = new ArrayList<Integer>();
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			adj[a].add(b);
			adj[b].add(a);
		}
		for (int i = 0; i < k; i++) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			adj2[a].add(b);
			adj2[b].add(a);
		}
		rnk = 0;
		for (int i = 0; i < n; i++) {
			rnk++;
			if (!vis[i]) {
				dst = dfs(i);
				dfsD(i);
			}
		}
		for (int i = 0; i < n; i++) {
			int ans=dist[i]-adj[i].size()-1;
			for(int v:adj2[i])
				ans-=(rank[v]==rank[i])?1:0;
			pw.print(ans+" ");
		}
		pw.close();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException, IOException {
			return br.ready();
		}

	}
}