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 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
const int MX = 200005;
vector<int> to[MX];
vector<int> to2[MX]; 
vector<int> group[MX];
vector<int> group2(MX);
void dfs(int v, int i,int p = -1) {
	group[i].push_back(v);
	group2[v] = i;
	for (int u : to[v]) {
		if (u == p) continue;
		if (0 == group2[u]) {
			dfs(u, i, v);
		}
	}
	//cout <<v<<" "<<in[v]<<" "<<out[v]<<endl;
	return;
}

int main() {
	int N, M, K;
	cin >>N>>M>> K;
	vector<pair<int,int>>P;
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		to[a].push_back(b);
		to[b].push_back(a); 
		P.push_back(make_pair(a, b));
	}	
	for (int i = 0; i < K; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		to2[a].push_back(b);
		to2[b].push_back(a);
		P.push_back(make_pair(a, b));
	}
	int num = 1;
	for (int n = 0;n < N;++n) {
		if (0 == group2[n]) {
			dfs(n, num);
			num++;
		}
	}
	vector<int> Ans(N);
	for (int i = 1;i<num;++i) {
		for (int j = 0; j < group[i].size();++j) {
			Ans[group[i][j]] = group[i].size()-1;
		}
	}
	for (int i = 0;i<P.size();++i) {
		if (group2[P[i].first] == group2[P[i].second]) {
			Ans[P[i].first]--;
			Ans[P[i].second]--;
		}
	}


	for (int i = 0; i < N; ++i) {
		if (0 != i) {
			cout << " ";
		}
		cout << Ans[i];
	}
	return 0;
}