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::io::*;
use std::str::FromStr;
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}
#[allow(dead_code)]
struct Main {
    n: usize,
    k: usize,
    f_graph: Vec<Vec<usize>>,
    labels: Vec<Option<usize>>,
    label_cnts: Vec<usize>,
    b_pairs: Vec<Option<(usize, usize)>>,
    visited: Vec<bool>,
}
#[allow(dead_code)]
impl Main {
    fn dfs(&mut self, i: usize, label: usize) {
        self.visited[i] = true;
        self.labels[i] = Some(label);
        self.label_cnts[label] += 1;
        for &new_i in self.f_graph[i].clone().iter() {
            if !self.visited[new_i] {
                self.dfs(new_i, label);
            }
        }
    }
    fn invoke(&mut self) {
        for i in 0..self.n {
            if !self.visited[i] {
                self.label_cnts.push(0);
                let label_i = self.label_cnts.len() - 1;
                self.dfs(i, label_i);
            }
        }
        let mut b_cnts: Vec<usize> = vec![0; self.n];
        for &b_pair in self.b_pairs.iter() {
            let a = b_pair.unwrap().0;
            let b = b_pair.unwrap().1;
            if self.labels[a] == self.labels[b] {
                b_cnts[a] += 1;
                b_cnts[b] += 1;
            }
        }
        for i in 0..self.n {
            let label = self.labels[i].unwrap();
            let res = self.label_cnts[label] - 1 - self.f_graph[i].len() - b_cnts[i];
            println!("{}", res);
        }
    }
}
fn main() {
    let n: usize = read();
    let m: usize = read();
    let k: usize = read();
    let mut f_graph: Vec<Vec<usize>> = vec![vec![]; n];
    for _ in 0..m {
        let a: usize = read::<usize>() - 1;
        let b: usize = read::<usize>() - 1;
        f_graph[a].push(b);
        f_graph[b].push(a);
    }
    let mut b_pairs: Vec<Option<(usize, usize)>> = vec![None; k];
    for i in 0..k {
        let a: usize = read::<usize>() - 1;
        let b: usize = read::<usize>() - 1;
        b_pairs[i] = Some((a, b));
    }
    Main {
        n: n,
        k: k,
        f_graph: f_graph,
        b_pairs: b_pairs,
        labels: vec![None; n],
        label_cnts: vec![],
        visited: vec![false; n],
    }.invoke();
}