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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Linq;

class D
{
	static int[] Read() => Console.ReadLine().Split().Select(int.Parse).ToArray();
	static void Main()
	{
		var h = Read();
		var n = h[0];
		var ab = new int[h[1]].Select(_ => Read()).ToArray();
		var cd = new int[h[2]].Select(_ => Read()).ToArray();

		var uf = new UF(n + 1);
		foreach (var x in ab)
			uf.Unite(x[0], x[1]);

		var roots = Enumerable.Range(0, n + 1).GroupBy(uf.GetRoot).ToDictionary(g => g.Key, g => g.Count());
		var c = Enumerable.Range(0, n + 1).Select(uf.GetRoot).Select(x => roots[x] - 1).ToArray();

		foreach (var x in ab)
		{
			c[x[0]]--;
			c[x[1]]--;
		}
		foreach (var x in cd)
		{
			if (!uf.AreUnited(x[0], x[1])) continue;
			c[x[0]]--;
			c[x[1]]--;
		}
		Console.WriteLine(string.Join(" ", c.Skip(1)));
	}
}

class UF
{
	int[] p;
	public UF(int n) { p = Enumerable.Range(0, n).ToArray(); }

	public void Unite(int a, int b) { if (!AreUnited(a, b)) p[p[b]] = p[a]; }
	public bool AreUnited(int a, int b) => GetRoot(a) == GetRoot(b);
	public int GetRoot(int a) => p[a] == a ? a : p[a] = GetRoot(p[a]);
}