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.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        
        UnionFind uf = new UnionFind(n);
        
        List<Integer>[] to = new List[n];
        for (int i = 0; i < n; i++) {
            to[i] = new ArrayList<Integer>();
        }
        
        for (int i = 0; i < m; i++) {
            int p = sc.nextInt() - 1;
            int q = sc.nextInt() - 1;
            if (p > q) {
                int tmp = p;
                p = q;
                q = tmp;
            }
            uf.unite(p, q);
            
            to[p].add(q);
            to[q].add(p);
        }
        
        List<Integer>[] to2 = new List[n];
        for (int i = 0; i < n; i++) {
            to2[i] = new ArrayList<Integer>();
        }
        
        for (int i = 0; i < k; i++) {
            int p = sc.nextInt() - 1;
            int q = sc.nextInt() - 1;
            
            to2[p].add(q);
            to2[q].add(p);
        }
        
        for (int i = 0; i < n; i++) {
            int count = uf.size[uf.root(i)];
            count--;
            count -= to[i].size();
            // System.out.println(count + "a");
            
            
            for (Integer x : to2[i]) {
                if (uf.same(x, i)) {
                    count--;
                }
            }
            
            System.out.print(count);
            if (i != n - 1) {
                System.out.print(" ");
            } else {
                System.out.println();
            }
        }
    }
}

class UnionFind {
    int[] par;  // 自身の親
    int[] size; // 自身の属するグループの要素数
    int connectedComponent;

    UnionFind(int n) {
        par = new int[n];
        for (int i = 0; i < n; i++) {
            par[i] = i;
        }
        
        size = new int[n];
        Arrays.fill(size, 1);
        connectedComponent = n;
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
        // 1-2-4の場合、4の親=2を4の根=1に繋ぎ変えた上で親を返す
        // ↑根まで辿る深さを減らす為
        // return a = b;とは、aにbを代入し、aを返すという意味
    }

    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        if (rx != ry) {
            par[ry] = rx;
            
            int resize = size[rx] + size[ry];
            size[rx] = resize;
            size[ry] = resize;
            connectedComponent--;
        }
    }

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