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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int p[100005], count[100005];
int root(int n)
{
	if (p[n] != n)
		p[n] = root(p[n]);
	return p[n];
}
void uni(int x, int y)
{
	x = root(x);
	y = root(y);
	if (x != y)
	{
		p[x] = y;
		count[y] += count[x];
	}
	return;
}
int main()
{
	int n, m, k;
	scanf("%d %d %d", &n, &m, &k);
	int i;
	int a[100005], b[100005];
	for (i = 0; i < m; i++)
	{
		scanf("%d %d", &a[i], &b[i]);
		a[i]--;
		b[i]--;
	}
	int c[100005], d[100005];
	for (i = 0; i < k; i++)
	{
		scanf("%d %d", &c[i], &d[i]);
		c[i]--;
		d[i]--;
	}
	for (i = 0; i < n; i++)
	{
		p[i] = i;
		count[i] = 1;
	}
	for (i = 0; i < m; i++)
		uni(a[i], b[i]);
	int e[100005];
	for (i = 0; i < n; i++)
		e[i] = 1;
	for (i = 0; i < m; i++)
	{
		if (root(a[i]) == root(b[i]))
		{
			e[a[i]]++;
			e[b[i]]++;
		}
	}
	for (i = 0; i < k; i++)
	{
		if (root(c[i]) == root(d[i]))
		{
			e[c[i]]++;
			e[d[i]]++;
		}
	}
	for (i = 0; i < n - 1; i++)
		printf("%d ", count[root(i)] - e[i]);
	printf("%d\n", count[root(n - 1)] - e[i]);
	return 0;
}