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>
using namespace std;

#define int long long int
#define ff first
#define ss second
#define pb push_back


int bexp(int a, int b)
{
	int res = 1;

	while (b)
	{
		if (b & 1) res = (res * a);

		a = (a * a); 
		b >>= 1;
	}
	return res;
}

int gcd(int a, int b)
{
	if (b == 0)
		return a;
	return gcd(b, a % b);
}

#define N 100010
#define MOD 1000000007

int visited[N];

vector<int> g1[N], g2[N];
vector<int> component;

void dfs(int v)
{
	visited[v] = 1;
	component.pb(v);

	for (auto e : g1[v])
	{
		if (!visited[e])
			dfs(e);
	}
}

int32_t main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.precision(10);

	int n, m, k;
	cin >> n >> m >> k;
	int a, b;
	for (int i = 0; i < m; ++i)
	{
		cin >> a >> b;
		g1[a].pb(b);
		g1[b].pb(a);
	}

	for (int i = 0; i < k; ++i)
	{
		cin >> a >> b;
		g2[a].pb(b);
		g2[b].pb(a);
	}

	int ans[n + 1] = {};
	for (int i = 0; i < n; ++i)
	{
		if (!visited[i + 1])
			dfs(i + 1);
		else
			continue;
		int no = component.size();
		for (auto e : component)
		{
			int k1 = 0, k2 = 0;
			k1 = g1[e].size();

			for (auto e1 : g2[e])
			{
				// cout << e1 << ' ';
				if (visited[e1] == 1)
					k2++;
			}

			// cout << e << " " << k1 << " " << k2 << "\n";
			ans[e] = no - 1 - k1 - k2;
			
		}

		for (auto e : component)
			visited[e] = 2;
		component.clear();
	}

	for (int i = 1; i <= n; ++i)
		cout << ans[i] << " ";
	return 0;
}