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;
using ll = long long;

ll dfs(ll node, vector<ll>& fg, const unordered_multimap<ll, ll>& ABm, ll group_no)
{
    if (fg[node] != 0) {
        return 0;
    }
    fg[node] = group_no;

    ll num = 1;
    auto r = ABm.equal_range(node);
    for (auto r = ABm.equal_range(node); r.first != r.second; r.first++) {
        num += dfs(r.first->second, fg, ABm, group_no);
    }
    return num;
}

void main_()
{
    ll N, M, K;
    cin >> N >> M >> K;
    unordered_multimap<ll, ll> ABm;
    for (ll i = 0; i < M; i++) {
        ll A, B;
        cin >> A >> B;
        ABm.emplace(A, B);
        ABm.emplace(B, A);
    }
    unordered_multimap<ll, ll> CDm;
    for (ll i = 0; i < K; i++) {
        ll C, D;
        cin >> C >> D;
        CDm.emplace(C, D);
        CDm.emplace(D, C);
    }

    vector<ll> fg(N + 1); // 1-origin

    unordered_map<ll, ll> group_num_m;
    for (ll i = 1; i <= N; i++) {
        if (fg[i] == 0) {
            auto num = dfs(i, fg, ABm, i);
            group_num_m[i] = num;
        }
    }

    for (ll i = 1; i <= N; i++) {
        ll ans = group_num_m[fg[i]] - 1; // 自分を含む友達グループから自分を引いた数
        auto it = ABm.equal_range(i);
        ans -= distance(it.first, it.second); // 自分と直接友達関係を除く

        for (auto it = CDm.equal_range(i); it.first != it.second; it.first++) {
            auto block_i = it.first->second;
            if (fg[i] == fg[block_i]) {
                // ブロック関係の相手が友達グループに含まれていたらその分を除く
                ans--;
            }
        }
        cout << ans << " ";
    }
    cout << "\n";
}
int main()
{
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);
    main_();
    return 0;
}