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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
 
#define mem(dp,a)           memset(dp,a,sizeof dp)
#define ninja               ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define pb(x)               push_back(x)
#define mp(x,y)             make_pair(x,y)
#define all(c)              c.begin(), c.end()
#define ff                  first
#define ss                  second
#define nl                  '\n'
#define inf                 1e9+7

vector<int> adj_frnd[100003];
vector<int> adj_blk[100003];

bool vis[100003];
bool reach;

void dfs(int s, int t){
	vis[s]=1;
	queue<int> q;
	q.push(s);
	while(!q.empty())
	{
		int p=q.front();
		if(p==t){
			reach=1; return;
		}
		for(int i=0;i<adj_frnd[p].size();i++){
			if(!vis[adj_frnd[p][i]]){
				vis[adj_frnd[p][i]]=1;
				q.push(adj_frnd[p][i]);
			}
		}
		q.pop();
	}
	
}
int main()
{
	ninja
	int n,m,k;
	cin>>n>>m>>k;
	int a,b;
	for(int i=0;i<m;i++){
		cin>>a>>b;
		adj_frnd[a].pb(b);
		adj_frnd[b].pb(a);
	}
	for(int i=0;i<k;i++){
		cin>>a>>b;
		adj_blk[a].pb(b);
		adj_blk[b].pb(a);
	}

	vector<int> v(n+1,0);

	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(i!=j){
				bool f=0;
				// search adj_frnd...
				for(auto l:adj_frnd[i]){
					if(l==j){
						f=1; break;
					}
				}
				if(f)continue;
				f=0;
				for(auto l:adj_blk[i]){
					if(l==j){
						f=1; break;
					}
				}
				if(f)continue;
				// chk reachability...

				reach=0;mem(vis,0);
				dfs(i,j);
				if(reach)
					v[i]++;
			}
		}
	}

	for(int i=1;i<=n;i++)
		cout<<v[i]<<" ";
}