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 (1.42.0) (judge verdict: Time Limit Exceeded)

#![allow(
  non_snake_case,
  unused_variables,
  unused_assignments,
  unused_mut,
  unused_imports,
  dead_code,
  unused_macros
)]
//use proconio::fastout;
use proconio::input;
use proconio::marker::*;
//use std::collections::HashSet;
use petgraph::unionfind::UnionFind;
use std::cmp::*;
use std::collections::*;
macro_rules! debug {
  ($($a:expr),* $(,)*) => {
      #[cfg(debug_assertions)]
      eprintln!(concat!($("| ", stringify!($a), "={:?} "),*, "|"), $(&$a),*);
  };
}
fn main() {
  input! {
     N:usize, M:usize, K:usize,
     AB:[(Usize1, Usize1);M],
     CD:[(Usize1, Usize1);K]
  }

  let mut t: UnionFind<usize> = UnionFind::new(N);
  for &(a, b) in AB.iter() {
    t.union(a, b);
  }
  let mut block: HashMap<usize, Vec<usize>> = HashMap::new();
  for &(c, d) in CD.iter() {
    (*block.entry(c).or_insert(vec![])).push(d);
    (*block.entry(d).or_insert(vec![])).push(c);
  }
  //let mut friend: HashMap<usize, Vec<usize>> = HashMap::new();
  for &(c, d) in AB.iter() {
    (*block.entry(c).or_insert(vec![])).push(d);
    (*block.entry(d).or_insert(vec![])).push(c);
  }
  for i in 0..N {
    let mut ans = 0;
    let mut bs = &Vec::new();
    match block.get(&i) {
      Some(x) => bs = x,
      None => (),
    };
    let top = t.find(i);
    debug!(i, top, bs,);
    for j in 0..N {
      if i != j && top == t.find(j) && bs.into_iter().find(|&&v| v == j) == None {
        ans += 1;
      }
    }
    print!("{} ", ans);
  }
}