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: Accepted)

use std::io::Read;
use std::collections::{HashMap, VecDeque};
use itertools::Itertools;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let m: usize = iterator.next().unwrap().parse().unwrap();
    let k: usize = iterator.next().unwrap().parse().unwrap();
    let mut graph = vec![Vec::new(); n];
    for _ in 0..m {
        let ai: usize = iterator.next().unwrap().parse().unwrap();
        let bi: usize = iterator.next().unwrap().parse().unwrap();
        graph[ai - 1].push(bi - 1);
        graph[bi - 1].push(ai - 1);
    }
    let mut blocks: Vec<(usize, usize)> = Vec::new();
    for _ in 0..k {
        let ci: usize = iterator.next().unwrap().parse().unwrap();
        let di: usize = iterator.next().unwrap().parse().unwrap();
        blocks.push((ci - 1, di - 1));
    }

    let mut colors: Vec<Option<usize>> = vec![None; n];
    let mut color: usize = 0;
    let mut counts: HashMap<usize, usize> = HashMap::new();
    for i in 0..n {
        if colors[i].is_some() {
            continue;
        }
        let mut q: VecDeque<usize> = VecDeque::new();
        q.push_back(i);
        colors[i] = Some(color);
        *counts.entry(color).or_insert(0) += 1;
        while let Some(u) = q.pop_front() {
            for &v in &graph[u] {
                if colors[v].is_some() {
                    continue;
                }
                q.push_back(v);
                colors[v] = Some(color);
                *counts.entry(color).or_insert(0) += 1;
            }
        }
        color += 1;
    }

    let mut result = vec![0; n];
    for i in 0..n {
        result[i] = *counts.get(&colors[i].unwrap()).unwrap() - graph[i].len() - 1;
    }

    for b in blocks {
        if colors[b.0] == colors[b.1] {
            result[b.0] -= 1;
            result[b.1] -= 1;
        }
    }

    result.iter().map(|it| it.to_string()).join(" ")
}