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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

#define N 100300
typedef struct {
	int apper;
	int group;
}node_t;

typedef struct {
	int member[N];
	int member_n;
}Group_t;

node_t node[N] = { 0 };
int group_n = 0;

int main() {
	int n, m, k;
	int a1[N] = { 0 }, b1[N] = { 0 };
	int c1[N] = { 0 }, d1[N] = { 0 };
	int min, max;
	int a, b, c, d;
	Group_t Group[N] = { 0 };

	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < m; i++) scanf("%d%d", &a1[i], &b1[i]);
	for (int i = 0; i < k; i++) scanf("%d%d", &c1[i], &d1[i]);
	for (int i = 0; i < m; i++) {
		a = a1[i], b = b1[i];
		node[a].apper++, node[b].apper++;
		if (node[a].group == 0 && node[b].group == 0) {
			group_n++, node[a].group = group_n, node[b].group = group_n;
			Group[group_n].member[0] = a;
			Group[group_n].member[1] = b;
			Group[group_n].member_n += 2;
		}
		else if (node[a].group != 0 && node[b].group == 0) {
			node[b].group = node[a].group;
			Group[node[a].group].member[Group[node[a].group].member_n] = b;
			Group[node[a].group].member_n++;
		}
		else if (node[a].group == 0 && node[b].group != 0) {
			node[a].group = node[b].group;
			Group[node[b].group].member[Group[node[b].group].member_n] = a;
			Group[node[b].group].member_n++;
		}
		else if (node[a].group != node[b].group) {
			if (Group[node[a].group].member_n < Group[node[b].group].member_n)
				min = node[a].group, max = node[b].group;
			else min = node[b].group, max = node[a].group;

			for (int j = 0; j < Group[min].member_n; j++) {
				Group[max].member[Group[max].member_n + j] = Group[min].member[j];
			}
			for (int p = 0; p < Group[min].member_n; p++) {
				node[Group[min].member[p]].group = max;
			}

			Group[max].member_n += Group[min].member_n;
		}
	}
	for (int i = 0; i < k; i++) {
		c = c1[i], d = d1[i];
		if (node[c].group != 0 || node[d].group != 0) {
			if (node[c].group == node[d].group)
				node[c].apper++, node[d].apper++;
		}

	}
	for (int i = 1; i <= n; i++) {
		if (node[i].group)
			printf("%d ", Group[node[i].group].member_n - node[i].apper - 1);
		else
			printf("0 ");
	}
	return 0;
}