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)

use proconio::input;
use std::collections::HashMap;
 
fn main() {
  input! {
    N: usize,
    M: usize,
    K: usize,
    A: [(usize, usize);M],
    B: [(usize, usize);K],
  }
  
  let mut group_num = 1;
  let mut map: HashMap<usize, usize> = HashMap::new();
  let mut connect_map: HashMap<usize, usize> = HashMap::new();
  for i in 1..=N {
    if map.get(&i).is_none() {
      let mut stack: Vec<usize> = vec![i];
      let mut count = 0;
      while 0 < stack.len() {
        let v = stack.pop().unwrap();
        for a in A.iter() {
          if a.0 == v && map.get(&a.1).is_none() {
            map.insert(a.1, group_num);
            count += 1;
            stack.push(a.1);
          }
          
          if a.1 == v && map.get(&a.0).is_none() {
            map.insert(a.0, group_num);
            count += 1;
            stack.push(a.0);
          }
        }
      }
      connect_map.insert(group_num, count);
      group_num += 1;
    }
  }

  let mut friends: Vec<usize> = vec![0;N+1];
  for a in A.iter() {
    friends[a.0] += 1;
    friends[a.1] += 1;
  }
  
  let mut blocks: Vec<usize> = vec![0;N+1];
  for b in B.iter() {
    if map.get(&b.0) == map.get(&b.1) {
      blocks[b.0] += 1;
      blocks[b.1] += 1;
    }
  }

  let mut result: Vec<usize> = vec![0;N];
  for i in 1..=N {
    if let Some(v) = map.get(&i) {
      let cv = connect_map.get(&v).unwrap();
      result[i-1] = cv - friends[i] - blocks[i] - 1;
    } else {
      result[i-1] = 0;
    }
  }
  
  println!("{}", result
    .into_iter()
    .map(|v| v.to_string())
    .collect::<Vec<String>>()
    .join(" ")
  );
}