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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

using Graph = vector<set<int>>;
using Seen = set<int>;

void dfs(Seen &seen, Graph G, int v)
{
    seen.insert(v);
    for (int f : G[v])
    {
        if (seen.count(f))
        {
            continue;
        }
        dfs(seen, G, f);
    }
}

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

    Seen seen;
    Graph frnd(N + 1);
    Graph block(N + 1);

    for (int i = 0; i < M; i++)
    {
        int a, b;
        cin >> a >> b;
        frnd[a].insert(b);
        frnd[b].insert(a);
    }
    for (int i = 0; i < K; i++)
    {
        int c, d;
        cin >> c >> d;
        block[c].insert(d);
        block[d].insert(c);
    }

    // cout << "Friends" << endl;
    // for (int i = 1; i <= N; i++)
    // {
    //     for (int j : frnd[i])
    //     {
    //         cout << i << "->" << j << endl;
    //     }
    // }

    // cout << "Blocks" << endl;
    // for (int i = 1; i <= N; i++)
    // {
    //     for (int j : block[i])
    //     {
    //         cout << i << "->" << j << endl;
    //     }
    // }

    // Main
    for (int i = 1; i <= N; i++)
    {
        seen.clear();

        dfs(seen, frnd, i);

        // for (int f : seen)
        // {
        //     cout << i << " -> " << f << endl;
        // }

        for (int b : block[i])
        {
            seen.erase(b);
        }

        // for (int f : seen)
        // {
        //     cout << "After block deletetion, " << i << " -> " << f << endl;
        // }

        for (int f : frnd[i])
        {
            seen.erase(f);
        }

        // for (int f : seen)
        // {
        //     cout << "After friend deletetion, " << i << " -> " << f << endl;
        // }

        cout << seen.size() - 1 << endl;
    }
}