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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num::clamp;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: usize, m: usize, k: usize,
        ab: [(Usize1, Usize1); m],
        cd: [(Usize1, Usize1); k],
    }
    
    let mut friend = vec![Vec::new(); n];
    let mut uft = UFT::new(n);
    for (a, b) in ab {
        friend[a].push(b);
        friend[b].push(a);
        uft.merge(a, b);
    }
    let mut block = vec![Vec::new(); n];
    for (c, d) in cd {
        block[c].push(d);
        block[d].push(c);
    }
    let mut chains = HashMap::new();
    for i in 0..n {
        let root = uft.root(i);
        *chains.entry(root).or_insert(0) += 1;
    }

    for i in 0..n {
        let root = uft.root(i);
        let chain = chains.get(&root).unwrap();
        let mut ans = chain - 1 - friend[i].len();
        for &bl in &block[i] {
            if uft.root(bl) == root {
                ans -= 1;
            }
        }
        println!("{}", ans);
    }
}

#[allow(dead_code)]
/// Union Find Tree
pub struct UFT {
    pub par: Vec<usize>,
    pub rank: Vec<usize>,
}
impl UFT {
    #[allow(dead_code)]
    pub fn new(n: usize) -> Self {
        UFT {
            par: (0..n).collect(),
            rank: vec![0; n],
        }
    }
    #[allow(dead_code)]
    pub fn root(&mut self, x: usize) -> usize {
        if self.par[x] == x {
            x
        } else {
            let p = self.par[x];
            let pp = self.root(p);
            self.par[x] = pp;
            pp
        }
    }
    #[allow(dead_code)]
    pub fn merge(&mut self, x: usize, y: usize) {
        let x = self.root(x);
        let y = self.root(y);
        if x == y {
            return;
        }
        if self.rank[x] < self.rank[y] {
            self.par[x] = y;
        } else {
            self.par[y] = x;
            if self.rank[x] == self.rank[y] {
                self.rank[x] += 1;
            }
        }
    }
}
*/

fn main() {
    let exe = "/tmp/bin4C4F0E54";
    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>";