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::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let m: usize = iterator.next().unwrap().parse().unwrap();
    let k: usize = iterator.next().unwrap().parse().unwrap();

    let mut friends: Vec<Vec<usize>> = vec![Vec::new(); n];
    for _i in 0..m {
        let a: usize = iterator.next().unwrap().parse().unwrap();
        let b: usize = iterator.next().unwrap().parse().unwrap();
        friends[a - 1].push(b - 1);
        friends[b - 1].push(a - 1);
    }


    let mut blocks: Vec<Vec<usize>> = vec![Vec::new(); n];
    for _i in 0..k {
        let c: usize = iterator.next().unwrap().parse().unwrap();
        let d: usize = iterator.next().unwrap().parse().unwrap();
        blocks[c - 1].push(d - 1);
        blocks[d - 1].push(c - 1);
    }


    let mut colors: Vec<usize> = vec![0; n];
    assign_color(n, &mut colors, &friends);

    let mut ans = String::new();
    for i in 0..n {
        let mut count = colors.iter().filter(|&&x| x == colors[i]).count();
        count -= 1;
        count -= friends[i].len();
        count -= blocks[i].iter().filter(|&&x| colors[x] == colors[i]).count();
        ans = format!("{} {}", ans, count);
    }

    return ans.trim().to_string();
}

fn assign_color(n: usize, colors: &mut Vec<usize>, friends: &Vec<Vec<usize>>) {
    let mut id: usize = 1;
    for i in 0..n {
        if colors[i] == 0 {
            dfs(i, id, colors, friends);
            id += 1;
        }
    }
}

fn dfs(r: usize, c: usize, colors: &mut Vec<usize>, friends: &Vec<Vec<usize>>) {
    let mut s: Vec<usize> = Vec::new();
    s.push(r);
    colors[r] = c;
    while !s.is_empty() {
        let u = s.pop().unwrap();
        for i in 0..friends[u].len() {
            let v = friends[u][i];
            if colors[v] == 0 {
                colors[v] = c;
                s.push(v);
            }
        }
    }
}