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: Time Limit Exceeded)

use std::collections::HashSet;

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

    let mut friends: Vec<HashSet<usize>> = vec![HashSet::new(); n+1];
    for _ in 0..m {
        let ab = read_vec::<usize>();
        let a = ab[0];
        let b = ab[1];

        friends[a].insert(b);
        friends[b].insert(a);
    }
    let mut blocks: Vec<HashSet<usize>> = vec![HashSet::new(); n+1];
    for _ in 0..k {
        let ab = read_vec::<usize>();
        let a = ab[0];
        let b = ab[1];

        blocks[a].insert(b);
        blocks[b].insert(a);
    }

    let mut ret = vec![0; n];
    for i in 1..n+1 {
        let mut done = vec![false; n+1];
        let cnt = dfs(i, i, &mut done, &friends, &blocks);
        ret[i-1] = cnt;
    }

    print!("{}", ret[0]);
    for i in 1..n {
        print!(" ");
        print!("{}", ret[i]);
    }
    println!("");
}

fn dfs(i: usize, base: usize, done: &mut Vec<bool>, friends: &Vec<HashSet<usize>>, blocks: &Vec<HashSet<usize>>) -> usize {
    if done[i] {
        return 0
    }
    done[i] = true;

    let mut cnt = 0;
    for friend in friends[i].iter() {
        cnt += dfs(*friend, base, done, friends, blocks);
    }

    if friends[base].contains(&i) || blocks[base].contains(&i) {
        return cnt
    }
    if i == base {
        return cnt
    }

    return cnt + 1
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}