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 (1.15.1) (judge verdict: Accepted)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::{fastout, input};
use std::collections::HashSet;

#[derive(Debug, Clone)]
struct Node {
    f: Vec<usize>,
    b: Vec<usize>,
    c: bool,
}

fn trace(nodes: &mut [Node], idx: usize, chain: &mut HashSet<usize>) {
    chain.insert(idx);
    nodes[idx].c = true;

    for f in nodes[idx].f.clone() {
        if nodes[f].c {
            continue;
        }

        trace(nodes, f, chain);
    }
}

#[fastout]
fn main() {
    input! {in_n: usize, in_m: usize, in_k: usize};
    input! {in_ab: [[usize; 2]; in_m], in_cd: [[usize; 2]; in_k]};
    let mut nodes = vec![
        Node {
            f: vec![],
            b: vec![],
            c: false,
        };
        in_n
    ];

    for ab in &in_ab {
        let a = ab[0] - 1;
        let b = ab[1] - 1;

        nodes[a].f.push(b);
        nodes[b].f.push(a);
    }

    for cd in &in_cd {
        let c = cd[0] - 1;
        let d = cd[1] - 1;

        nodes[c].b.push(d);
        nodes[d].b.push(c);
    }

    let mut count = vec![0; in_n];

    for i in 0..nodes.len() {
        if nodes[i].c {
            continue;
        }

        let mut chain = HashSet::new();

        trace(&mut nodes, i, &mut chain);

        for j in &chain {
            let mut c = chain.len() - 1 - nodes[*j].f.len();

            for b in &nodes[*j].b {
                if chain.contains(b) {
                    c -= 1;
                }
            }

            count[*j] = c;
        }
    }

    println!(
        "{}",
        count
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(" ")
    );
}
*/

fn main() {
    let exe = "/tmp/binA5E61796";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";