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>
#include <stdlib.h>

#define N 100005

typedef struct {
	int parent;
	int appaer;
	int member;
	int root;
	int visit;
}node_t;

node_t node[N] = { 0 };

void joint(int a, int b) {
	int a_root = root(a), b_root = root(b);

	if (a_root != b_root) {
		if (node[a_root].member >= node[b_root].member){
			node[b_root].parent = a_root;
			node[a_root].member += node[b_root].member;
			node[b_root].member = 0;
		}
		else {
			node[a_root].parent = b_root;
			node[b_root].member += node[a_root].member;
			node[a_root].member = 0;
		}
	}

}

int root(int node_n) {
	if (node[node_n].parent == 0)return node_n;
	else return root(node[node_n].parent);
}
int main(void) {
	int n, m, k;
	int a, b;
	int c, d;

	scanf("%d %d %d", &n, &m, &k);
	for (int i = 1; i <= n; i++)node[i].member = 1;
	for (int i = 0; i < m; i++) {
		scanf("%d %d", &a, &b);
		node[a].appaer++, node[b].appaer++;
			joint(a, b);
	}
	for (int i = 1; i <= n; i++)
		node[i].root = root(i);
	for (int i = 0; i < k; i++) {
		scanf("%d %d", &c, &d);
		if (node[c].root == node[d].root) {
			node[c].appaer++, node[d].appaer++;
		}
	}

	/*
	for (int i = 1; i <= n; i++) {
		printf("node[%d].root=%d\n", i, node[i].root);
	}
	for (int i = 1; i <= n; i++) {
		printf("node[%d].member=%d\n", i, node[i].member);
	}
	*/
	for (int i = 1; i <= n; i++) {
		if (node[node[i].root].member > 1)
			printf("%d ", node[node[i].root].member - node[i].appaer - 1);
		else printf("0 ");
	}
	return 0;
}