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>
#define ll long long
#define ld long double
#define endl "\n"
#define MOD 1000000007
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.precision(10);
    

    int n,m,k;
    cin >> n >> m >> k;   
    // friendship is a map implemented by adjacency list
    vector<int> friendship[n];
    for (int i=0;i<=m-1;i++)
    {
    	int a,b;
    	cin >> a >> b;
    	a--;
    	b--;
    	friendship[a].push_back(b);
    	friendship[b].push_back(a);
    }

    // blockship is implemented as a array of adjacency list
    vector<int> blockship[n];
    for (int i=0;i<=k-1;i++)
    {
    	int a,b;
    	cin >> a >> b;
    	a--;
    	b--;
    	blockship[a].push_back(b);
    	blockship[b].push_back(a);
    }

    for (int i=0;i<=n-1;i++)
    {
    	// counter for plausible friend candidates
    	// 1-possible
    	// 0-not possible
    	
    	int counter[n];
    	for (int j=0;j<=n-1;j++)
    	{
    		if (i!=j)
    		{
    			counter[j]=1;
    		}
    		else
    		{
    			counter[j]=0;
    		}
    	}

    	for (auto it=friendship[i].begin();it!=friendship[i].end();it++)
    	{
    		counter[*it]=0;
    	}

    	for (auto it=blockship[i].begin();it!=blockship[i].end();it++)
    	{
    		counter[*it]=0;
    	}

    	// we generate a set full of childs of arr[i] by performing a dfs on friendship
    	set<int> alpha;
    	stack<int> temp;

    	for (auto it=friendship[i].begin();it!=friendship[i].end();it++)
    	{
    		temp.push(*it);
    		alpha.insert(*it);
    	}

    	while (temp.size()!=0)
    	{
    		int q=temp.top();
    		temp.pop();
    		for (auto it=friendship[q].begin();it!=friendship[q].end();it++)
    		{
    			if (alpha.find(*it)==alpha.end())
    			{
    				// we did not found the element in alpha
    				// it is unvisited
    				alpha.insert(*it);
    				temp.push(*it);
    			}
    			// we found the element in alpha
    			// it is already visited so we do not do anything
    		}
    	}

    	// intersection between elements with counter[i]=1 and in alpha set is the answer
    	int ans=0;
    	for (int j=0;j<=n-1;j++)
    	{
    		if ((counter[j]==1)&&(alpha.find(j)!=alpha.end()))
    		{
    			ans++;
    		}
    	}
    	cout << ans << " ";
    }
    cout << endl;

    return 0;
}