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)

pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

#[derive(Debug)]
struct UnionFind {
    node: usize,
    parent: usize,
    friend: u32,
    block: u32,
    size: u32,
}

fn friend(a: usize, b: usize, uf: &mut Vec<UnionFind>) {
    uf[a].friend += 1;
    uf[b].friend += 1;
    let pa = parent(a, uf);
    let pb = parent(b, uf);
    if pa == pb {
        return;
    }
    if uf[pa].size < uf[pb].size {
        uf[pa].parent = pb;
        uf[pb].size += uf[pa].size;
    } else {
        uf[pb].parent = pa;
        uf[pa].size += uf[pb].size;
    }
}

fn block(a: usize, b: usize, uf: &mut Vec<UnionFind>) {
    let pa = parent(a, uf);
    let pb = parent(b, uf);
    if pa == pb {
        uf[a].block += 1;
        uf[b].block += 1;
    }
}

fn parent(a: usize, uf: &mut Vec<UnionFind>) -> usize {
    if uf[a].parent == 0 {
        uf[a].node
    } else {
        let p = parent(uf[a].parent, uf);
        uf[a].parent = p;
        p
    }
}

fn main() {
    let nmk: Vec<usize> = read_vec();
    let n = nmk[0];
    let m = nmk[1];
    let k = nmk[2];
    let abv: Vec<Vec<usize>> = read_vec2(m);
    let cdv: Vec<Vec<usize>> = read_vec2(k);

    let mut uf: Vec<UnionFind> = vec![];

    for i in 0..n + 1 {
        uf.push(UnionFind {
            node: i,
            parent: 0,
            friend: 0,
            block: 0,
            size: 1,
        });
    }

    for i in 0..m {
        let a = abv[i][0];
        let b = abv[i][1];
        friend(a, b, &mut uf);
    }

    for i in 0..k {
        let c = cdv[i][0];
        let d = cdv[i][1];
        block(c, d, &mut uf);
    }

    for i in 1..n + 1 {
        let p = parent(i, &mut uf);
        let c = uf[p].size - uf[i].friend - uf[i].block - 1;
        print!("{}", c);
        if i == n {
            println!("");
        } else {
            print!(" ");
        }
    }
}