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.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int m = scan.nextInt();
		int k = scan.nextInt();

		unionFind uf = new unionFind(n);

		int i,x,y;

		int[] f = new int[n + 1];
		int[] r = new int[n + 1];
		int[] rc = new int[n + 1];
		for(i = 0; i < m; i++) {
			x = scan.nextInt();
			y = scan.nextInt();

			f[x]++;
			f[y]++;

			uf.union(x, y);
		}

		for(i = 1; i <= n; i++) {
			r[i] = uf.root(i);
			rc[r[i]]++;
		}

		for(i = 0; i < k; i++) {
			x = scan.nextInt();
			y = scan.nextInt();

			if(uf.same(x, y)) {
				f[x]++;
				f[y]++;
			}
		}

		long temp;
		for(i = 1; i <= n; i++) {
			temp = rc[r[i]] - f[i] - 1;
			System.out.print(temp + " ");
		}
	}

}
class unionFind{
	int[] par;

	unionFind(int x){
		this.par = new int[x + 1];

		int i;
		for(i = 0; i <= x; i++)
			this.par[i] = i;
	}

	int root(int x) {
		if(this.par[x] == x)
			return x;

		return root(this.par[x]);
	}

	void union(int x, int y) {
		int rx = root(x);
		int ry = root(y);
		if(rx == ry)
			return;

		this.par[rx] = ry;
	}

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

		return rx == ry;
	}
}