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)

/**
 * Title: D - Friend Suggestions
 * Url: https://atcoder.jp/contests/abc157/tasks/abc157_d
 */
#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int MAX_N = 100000;

// 連結成分表作成
vector<int> dfs(int n, vector<vector<int>> G) {
    int i, j, vertex, neighbor;
    vector<int> comp(n+1, 0);
    vector<int> used(n+1, 0);
    stack<int> S;

    for(int i=1; i<=n; i++) {
        if(used[i]==0) {
            // dfs
            S.push(i);
            while(!S.empty()) {
                vertex = S.top();
                S.pop();
                used[vertex] = 1;
                comp[vertex] = i;
                for(j=0; j<(int) G[vertex].size(); j++) {
                    neighbor = G[vertex][j];    // 隣接している友人
                    if(used[neighbor]==0) S.push(neighbor);
                }
            }
        }
    }
    return comp;
}

int main() {
    int n, m, k; cin >> n >> m >> k;

    vector<vector<int>> friends(n+1);
    vector<vector<int>> nfriends(n+1);

    // 友達の登録
    for(int i=0; i<m; i++) {
        int ai, bi; cin >> ai >> bi;
        friends[ai].push_back(bi);
        friends[bi].push_back(ai);
    }

    // ブロック
    for(int i=0; i<k; i++) {
        int ci, di; cin >> ci >> di;
        nfriends[ci].push_back(di);
        nfriends[di].push_back(ci);
    }

    // 連結成分表
    auto comp = dfs(n, friends);

    // O(N)
    map<int, int> dic;
    for(auto e:comp) {
        dic[e]++;
    }

    vector<int> nfnum(n+1, 0);
    for(int i=1; i<=n; i++) {
        for(auto p:nfriends[i]) {
            if(comp[p] == comp[i]) nfnum[i]++;
        }
    }

    for(int i=1; i<=n; i++) {
        if(i==n) {
            cout << dic[comp[i]] - friends[i].size() - nfnum[i] - 1 << endl;
        } else {
            cout << dic[comp[i]] - friends[i].size() - nfnum[i] - 1 << " ";
        }
    }

    return 0;
}