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)

use std::mem::swap;
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

struct UnionFind {
    par: Vec<usize>,
    siz: Vec<usize>,
}

impl UnionFind {
    fn init(&mut self, n: usize) -> &mut UnionFind {
        for i in 0..n {
            self.par.push(i);
            self.siz.push(1);
        }
        self
    }

    fn find(&mut self, x: usize) -> usize {
        let ele = self.par[x];
        return if ele == x {
            x
        } else {
            self.par[x] = self.find(ele);
            self.par[x]
        };
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.find(x) == self.find(y)
    }

    fn unite(&mut self, x: usize, y: usize) -> bool {
        let mut x = self.find(x);
        let mut y = self.find(y);
        if x == y {
            return false;
        }
        if self.par[x] < self.par[y] {
            swap(&mut x, &mut y)
        }
        self.siz[x] += self.siz[y];
        self.par[y] = x;
        true
    }

    fn size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.siz[root]
    }
}

fn main() {
    input!(
    n: usize,
    m: usize,
    k: usize,
    f: [[usize; 2]; m],
    b: [[usize; 2]; k]
    );
    let mut u = UnionFind { par: vec![], siz: vec![] };
    let u = u.init(n);
    let mut fc = vec![0usize; n];
    for fr in f {
        u.unite(fr[0] - 1, fr[1] - 1);
        fc[fr[0] - 1] += 1;
        fc[fr[1] - 1] += 1;
    }
    for br in b {
        if u.same(br[0] - 1, br[1] - 1) {
            fc[br[0] - 1] += 1;
            fc[br[1] - 1] += 1;
        };
    }
    for i in 0..n {
        print!("{} ", u.size(i) - fc[i] - 1)
    }
}