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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let n: usize = scan(tokens);
    let mut sns = Sns::new(n);
    let m: usize = scan(tokens);
    let k: usize = scan(tokens);
    for _ in 0..m {
        let a: usize = scan(tokens);
        let b: usize = scan(tokens);
        sns.add_friend(a - 1, b - 1);
    }
    for _ in 0..k {
        let c: usize = scan(tokens);
        let d: usize = scan(tokens);
        sns.add_block(c - 1, d - 1);
    }
    sns.solve();
}

struct Sns {
    n: usize,
    friends: Vec<Vec<usize>>,
    blocks: Vec<Vec<usize>>,
    visited: Vec<bool>,
    root: Vec<usize>,
    count: Vec<usize>,
}

impl Sns {
    fn new(n: usize) -> Sns {
        Sns {
            n: n,
            friends: vec![Vec::new(); n],
            blocks: vec![Vec::new(); n],
            visited: vec![false; n],
            root: vec![0; n],
            count: vec![0; n],
        }
    }
    fn add_friend(&mut self, i: usize, j: usize) {
        self.friends[i].push(j);
        self.friends[j].push(i);
    }
    fn add_block(&mut self, i: usize, j: usize) {
        self.blocks[i].push(j);
        self.blocks[j].push(i);
    }
    fn solve(&mut self) {
        for i in 0..self.n {
            self.dfs(i, i);
        }
        for i in 0..self.n {
            let r = self.root[i];
            let mut cnt = self.count[r] - self.friends[i].len() - 1;
            for &j in &self.blocks[i] {
                if self.root[j] == r {
                    cnt -= 1;
                }
            }
            if i != 0 {
                print!(" ");
            }
            print!("{}", cnt);
        }
        println!("");
    }
    fn dfs(&mut self, i: usize, r: usize) {
        if self.visited[i] {
            return;
        }
        self.visited[i] = true;
        self.root[i] = r;
        self.count[r] += 1;
        for j in self.friends[i].clone() {
            self.dfs(j, r);
        }
    }
}