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:
/*
#[macro_use]
mod utils {
    macro_rules! input {
        ($($r:tt)*) => {
            let s = {
                use std::io::Read;
                let mut s = String::new();
                std::io::stdin().read_to_string(&mut s).unwrap();
                s
            };
            let mut iter = s.split_whitespace();
            input_inner!{iter, $($r)*}
        };
    }
    macro_rules! input_inner {
        ($iter:expr) => {};
        ($iter:expr, ) => {};

        ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
            let $var = read_value!($iter, $t);
            input_inner!{$iter $($r)*}
        };
        ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
            let mut $var = read_value!($iter, $t);
            input_inner!{$iter $($r)*}
        };
    }
    macro_rules! read_value {
        ($iter:expr, ( $($t:tt),* )) => {
            ( $(read_value!($iter, $t)),* )
        };
        ($iter:expr, [ $t:tt ; $len:expr ]) => {
            (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
        };
        ($iter:expr, chars) => {
            read_value!($iter, String).chars().collect::<Vec<char>>()
        };
        ($iter:expr, usize1) => {
            read_value!($iter, usize) - 1
        };
        ($iter:expr, $t:ty) => {
            $iter.next().unwrap().parse::<$t>().expect("Parse error")
        };
    }
}

struct UnionFind {
    n: Vec<(usize, usize)>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let mut uf = Self {
            n: Vec::with_capacity(n),
        };
        for i in 0..n {
            uf.n.push((i, 1));
        }
        uf
    }
    fn root(&mut self, i: usize) -> usize {
        if self.n[i].0 == i {
            return i;
        }
        let r = self.root(self.n[i].0);
        self.n[i].0 = r;
        r
    }
    fn size(&mut self, i: usize) -> usize {
        let r = self.root(i);
        self.n[r].1
    }
    fn unite(&mut self, i: usize, j: usize) {
        let i = self.root(i);
        let j = self.root(j);
        self.n[i].0 = j;
        self.n[j].1 += std::mem::replace(&mut self.n[i].1, 0);
    }
}

fn main() {
    input! {
        n:usize, m:usize, k:usize,
        ab:[(usize1, usize1); m],
        cd:[(usize1, usize1); k],
    }
    let mut uf = UnionFind::new(n);
    let mut ffblock = vec![vec![]; n];
    for (a, b) in ab {
        uf.unite(a, b);
        ffblock[a].push(b);
        ffblock[b].push(a);
    }
    for (c, d) in cd {
        ffblock[c].push(d);
        ffblock[d].push(c);
    }
    for i in 0..n {
        let mut r = 0;
        r += uf.size(i);
        for fb in &ffblock[i] {
            if uf.root(i) == uf.root(*fb) {
                r -= 1;
            }
        }
        print!("{} ", r - 1);
    }
}
*/

fn main() {
    let exe = "/tmp/bin0998522F";
    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>";