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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <boost/dynamic_bitset.hpp>
using namespace std;
using u32 = uint32_t;

void dfs(const vector<vector<u32>>& fs, shared_ptr<boost::dynamic_bitset<>>& ac,
map<u32, shared_ptr<boost::dynamic_bitset<>>>& mm, const u32 n) {
	
	(*ac).set(n);
	for(auto f : fs[n]) {
		if(!(*ac)[f]) {
			dfs(fs, ac, mm, f);
		}
	}
	mm[n] = ac;
}

int main() {
	u32 N, M, K;
	cin >> N >> M >> K;

	vector<vector<u32>> fs(N, vector<u32>());
	vector<vector<u32>> bs(N, vector<u32>());
	
	for(u32 i=0; i<M; i++) {
		u32 A, B;
		cin >> A >> B;
		fs[A-1].push_back(B-1);
		fs[B-1].push_back(A-1);
	}
	
	for(u32 i=0; i<K; i++) {
		u32 C, D;
		cin >> C >> D;
		bs[C-1].push_back(D-1);
		bs[D-1].push_back(C-1);
	}
	
	map<u32, shared_ptr<boost::dynamic_bitset<>>> mm;
	
	for(u32 i=0; i<N; i++) {
		auto itr = mm.find(i);
		boost::dynamic_bitset<> aci;
		
		if(mm.find(i) == mm.end()) {
			shared_ptr<boost::dynamic_bitset<>> ac(new boost::dynamic_bitset<>(N));
			dfs(fs, ac, mm, i);
			aci = *ac;
		} else {
			aci = *((*itr).second);
		}
		
		for(auto f : fs[i]) aci[f] = 0;
		for(auto b : bs[i]) aci[b] = 0;
		aci[i] = 0;
		cout << aci.count() << " ";
		mm.erase(i);
	}
	
	return 0;
}