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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;


class UnionFind
{
    /// <summary>
    /// 正の値は親nodeを示す。
    /// 負の値は、そのnodeが根であることと、そのnodeを根とする木のサイズ * (-1) を示す。
    /// </summary>
    public int[] Data;

    public UnionFind(int nNodes=0){
        Data = new int[nNodes];
        for(int i=0; i<nNodes; i++){
            Data[i] = -1;
        };
    }

    /// <summary>
    /// iNode番目のノードが属する木の根の番号を返す。
    /// </summary>
    int FindRoot(int iNode){
        if(Data[iNode] < 0){
            return iNode;
        }else{
            return Data[iNode] = FindRoot(Data[iNode]);
        }
    }

    /// <summary>
    /// iNodeが属する木のサイズを返す
    /// </summary>
    public int Size(int iNode){
        return -Data[FindRoot(iNode)];
    }

    
    /// <summary>
    /// 2つの木をくっつける。
    /// くっつけたらtrue, すでに同一の木であればfalseを返す。
    /// </summary>
    public bool Unite(int i1, int i2){
        i1 = FindRoot(i1);
        i2 = FindRoot(i2);
        if(i1 == i2){
            return false;
        }

        // 木が深くならないように (深くなるとFindの計算量がふえる)、
        // サイズが小さい木をサイズが大きい木へマージする
        int iLarge, iSmall;
        if(i1 > i2){
            // i1, i2 はサイズ * (-1)を表す！
            iLarge = i2;
            iSmall = i1;
        }else{
            iLarge = i1;
            iSmall = i2;
        }
        Data[iLarge] += Data[iSmall];
        Data[iSmall] = iLarge;
        return true;
    }

    /// <summary>
    /// 2つのノードが同一の木に属しているか否かを返す
    /// </summary>
    public bool HasSameRoot(int i1, int i2){
        return FindRoot(i1) == FindRoot(i2);
    }
}

class Program
{
    static void Main(string[] args)
    {
        string[] nmk = Console.ReadLine().Split(' ');
        int 
        N = int.Parse(nmk[0]),
        M = int.Parse(nmk[1]),
        K = int.Parse(nmk[2]);

        UnionFind uf = new UnionFind(N);
        int[] nFriends = new int[N];
        for(int i=0; i<M; i++){
            string[] ab = Console.ReadLine().Split(' ');
            int a = int.Parse(ab[0]) - 1, b = int.Parse(ab[1]) - 1;
            nFriends[a]++;
            nFriends[b]++;
            uf.Unite(a, b);
        }

        int[] nBlockedCands = new int[N];
        for(int i=0; i<K; i++){
            string[] ab = Console.ReadLine().Split(' ');
            int a = int.Parse(ab[0]) - 1, b = int.Parse(ab[1]) - 1;
            if(uf.HasSameRoot(a, b)){
                nBlockedCands[a]++;
                nBlockedCands[b]++;
            }
        }

        for(int i=0; i<N; i++){
            Console.Write(uf.Size(i) - nFriends[i] - nBlockedCands[i] - 1 /* self */);
            Console.Write(i == N-1 ? '\n' : ' ');
        }
    }
}