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)

// ABC 157-D
// 2020.3.1 bal4u

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

typedef long long ll;

#define gc() getchar_unlocked()
#define pc(c) putchar_unlocked(c)

int in() {   // 非負整数の入力
	int n = 0; int c;
	c = gc();
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

void out(int n) { // 非負整数の表示（出力）
	int i; char b[30];

	if (!n) pc('0');
	else {
		//		if (n < 0) pc('-'), n = -n;
		i = 0; while (n) b[i++] = n % 10 + '0', n /= 10;
		while (i--) pc(b[i]);
	}
}

#define MAX 100005

/* UNION-FIND library */
int id[MAX], size[MAX];
void init(int n) { int i; for (i = 0; i < n; i++) id[i] = i, size[i] = 1; }
int root(int i) { while (i != id[i]) id[i] = id[id[i]], i = id[i]; return i; }
int connected(int p, int q) { return root(p) == root(q); }
void unite(int p, int q) {
    int i = root(p), j = root(q); if (i == j) return;
    if (size[i] < size[j]) id[i] = j, size[j] += size[i]; else id[j] = i, size[i] += size[j];
}

int N; // グラフの頂点数
int hi[MAX], *to[MAX];
int x[MAX];

void getGraph1(int M) {
	int i, k, a, b, c;
	int *memo, sz;

	init(N+1);
	memo = malloc(M*2*sizeof(int));
	sz = 0; for (i = 0; i < M; ++i) {
		a = in(), b = in();
		unite(a, b);
		hi[a]++, hi[b]++;
		memo[sz++] = a, memo[sz++] = b;
	}
	for (i = 1; i <= N; ++i) if (hi[i]) {
		to[i] = malloc(hi[i]*sizeof(int));
	}
	memset(hi, 0, sizeof(hi));

	i = 0; while (i < sz) {
		a = memo[i++], b = memo[i++];
		k = hi[a]++, to[a][k] = b;
		k = hi[b]++, to[b][k] = a;
	}
	free(memo);
}

int main()
{
	int i, M, K;

	N = in(), M = in(), K = in();
	getGraph1(M);
	while (K--) {
		int c = in(), d = in();
		if (connected(c, d)) x[c]++, x[d]++;
	}
	out(size[root(1)]-1 - hi[1]-x[1]);
	for (i = 2; i <= N; ++i) {
		pc(' '), out(size[root(i)]-1 - hi[i]-x[i]);
	}
	pc('\n');
	return 0;
}