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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

fn main() {
    proconio::input! {
        n: usize,
        m: usize,
        k: usize,
        abs: [(usize, usize); m],
        cds: [(usize, usize); k],
    }
    let mut friends_map =
        std::collections::HashMap::<usize, std::collections::HashSet<usize>>::new();
    let mut blocks_map =
        std::collections::HashMap::<usize, std::collections::HashSet<usize>>::new();
    for i in 1..=n {
        friends_map.insert(i, std::collections::HashSet::new());
        blocks_map.insert(i, std::collections::HashSet::new());
    }
    for (a, b) in &abs {
        friends_map.entry(*a).or_default().insert(*b);
        friends_map.entry(*b).or_default().insert(*a);
    }
    for (c, d) in &cds {
        blocks_map.entry(*c).or_default().insert(*d);
        blocks_map.entry(*d).or_default().insert(*c);
    }
    let mut anss = vec![];
    for i in 1..=n {
        let mut count = 0usize;
        let blocks_of_i = blocks_map.get(&i).unwrap();
        let friends_of_i = friends_map.get(&i).unwrap();
        let mut deque = std::collections::VecDeque::new();
        deque.push_back(i);
        let mut used = std::collections::HashSet::new();
        while let Some(friend) = deque.pop_front() {
            if used.contains(&friend) {
                continue;
            }
            used.insert(friend);
            if friend != i && !blocks_of_i.contains(&friend) && !friends_of_i.contains(&friend) {
                count += 1;
            }
            if let Some(friends_of_friend) = friends_map.get(&friend) {
                for friend_of_friend in friends_of_friend {
                    if used.contains(friend_of_friend) {
                        continue;
                    }
                    deque.push_back(*friend_of_friend);
                }
            }
        }
        anss.push(count.to_string());
    }
    let ans = anss.join(" ");
    println!("{}", ans);
}