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)

#[allow(unused_imports)]
use std::cmp::{Ordering};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};

#[allow(unused_imports)]
use std::iter::FromIterator;

#[allow(unused_macros)]
macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::min($x, min!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($z: expr),+) => {{
        std::cmp::max($x, max!($($z),*))
    }}
}

#[allow(unused_macros)]
macro_rules! get { 
    ([$t: ty]) => { 
        { 
            let mut line = String::new(); 
            stdin().read_line(&mut line).unwrap(); 
            line.split_whitespace().map(|t|t.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    };
    ([$t: ty];$n: expr) => {
        (0..$n).map(|_|get!([$t])).collect::<Vec<_>>()
    };
    ($t: ty) => {
        {
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            line.trim().parse::<$t>().unwrap()
        }
    };
    ($($t: ty),*) => {
        { 
            let mut line = String::new();
            stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
    ($t: ty; $n: expr) => {
        (0..$n).map(|_|get!($t)).collect::<Vec<_>>()
    };
    ($($t: ty),*; $n: expr) => {
        (0..$n).map(|_|get!($($t),*)).collect::<Vec<_>>()
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

const BIG_STACK_SIZE: bool = true;

#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}

pub struct UnionFindSize {
    parent: Vec<usize>,
    size: Vec<usize>,
}
impl UnionFindSize {
#[allow(dead_code)]
    pub fn new(n: usize) -> Self {
        let mut parent = vec![0; n];
        for i in 0..n {
            parent[i] = i;
        }
        UnionFindSize {
            parent: parent,
            size: vec![1; n],
        }
    }
#[allow(dead_code)]
    pub fn root(&mut self, a: usize) -> usize {
        let par = self.parent[a];
        if par != a {
            self.parent[a] = self.root(par);
        }
        self.parent[a]
    }
#[allow(dead_code)]
    pub fn unite(&mut self, a: usize, b: usize) {
        let a = self.root(a);
        let b = self.root(b);
        if a == b {
            return;
        }
        if self.size[a] > self.size[b] {
            self.parent[b] = a;
            self.size[a] += self.size[b];
        } else {
            self.parent[a] = b;
            self.size[b] += self.size[a];
        }
    }
#[allow(dead_code)]
    pub fn is_same(&mut self, a: usize, b: usize) -> bool {
        self.root(a) == self.root(b)
    }
#[allow(dead_code)]
    pub fn size(&mut self, a: usize) -> usize {
        let par = self.root(a);
        self.size[par]
    }
}

fn solve() {
    let (n,m,k) = get!(usize, usize, usize);
    let ab = get!(usize, usize; m);
    let cd  = get!(usize, usize; k);

    let mut tree= UnionFindSize::new(n+1);
    let mut friend = vec![Vec::new();n+1];
    let mut block = vec![Vec::new(); n+1];
    for (a,b) in ab {
        tree.unite(a,b);
        friend[a].push(b);
        friend[b].push(a);
    }

    for (c,d) in cd {
        block[c].push(d);
        block[d].push(c);
    }

    for i in 1..n+1 {
        let mut count = 0;
        for &b in &block[i] {
            if tree.is_same(i,b) {
                count += 1;
            }
        }
        print!("{} ", tree.size(i)-friend[i].len()-count-1);
    }
    println!("");
}