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>

int N, M, K;

int tc[114514];
int* ts[114514];

int bc[114514];
int* bs[114514];

void ae(int* c, int** s, int t) {
	*s = realloc(*s, sizeof(int) * (*c + 1));
	if (*s == NULL) exit(2);
	(*s)[(*c)++] = t;
}

int tu[114514];
int bu[114514];

void ui(int* u) {
	int i;
	for (i = 0; i < 114514; i++) u[i] = -1;
}

int ur(int* u, int n) {
	if (u[n] < 0) return n;
	return u[n] = ur(u, u[n]);
}

int us(int* u, int n) {
	return -u[ur(u, n)];
}

void um(int* u, int n1, int n2) {
	int r1 = ur(u, n1);
	int r2 = ur(u, n2);
	if (r1 != r2) {
		int s1 = us(u, n1);
		int s2 = us(u, n2);
		if (s1 <= s2) {
			u[r2] += u[r1];
			u[r1] = r2;
		} else {
			u[r1] += u[r2];
			u[r2] = r1;
		}
	}
}

int main(void) {
	int i;
	if (scanf("%d%d%d", &N, &M, &K) != 3) return 1;
	ui(tu);
	ui(bu);
	for (i = 0; i < M; i++) {
		int A, B;
		if (scanf("%d%d", &A, &B) != 2) return 1;
		ae(&tc[A], &ts[A], B);
		ae(&tc[B], &ts[B], A);
		um(tu, A, B);
	}
	for (i = 0; i < K; i++) {
		int C, D;
		if (scanf("%d%d", &C, &D) != 2) return 1;
		ae(&bc[C], &bs[C], D);
		ae(&bc[D], &bs[D], C);
		um(bu, C, D);
	}

	for (i = 1; i <= N; i++) {
		int kaunto = us(tu, i) - 1 - tc[i];
		int j;
		int tr = ur(tu, i);
		for (j = 0; j < bc[i]; j++) {
			if (ur(tu, bs[i][j]) == tr) kaunto--;
		}
		printf(" %d" + (i == 1), kaunto);
	}
	putchar('\n');

	return 0;
}