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: Accepted)

use proconio::input;
use proconio::marker::Usize1;
use std::collections::BTreeSet;

fn main() {
    input! {
        n: usize, m: usize, k: usize,
        ab: [(Usize1,Usize1); m],
        cd: [(Usize1,Usize1); k]
    }

    let friend = {
        let mut friend = vec![Vec::new(); n];
        for &(a,b) in &ab {
            friend[a].push(b);
            friend[b].push(a);
        }
        friend
    };

    let block = {
        let mut block = vec![Vec::new(); n];
        for &(c,d) in &cd {
            block[c].push(d);
            block[d].push(c);
        }
        block
    };

    let ans = {
        let mut ans = vec![0; n];
        let mut arr = vec![false; n];
        for i in 0..n {
            if arr[i] {
                continue;
            } else {
                let mut stack = Vec::new();
                stack.push(i);
                let mut set = BTreeSet::new();
                while let Some(j) = stack.pop() {
                    arr[j] = true;
                    set.insert(j);
                    for &k in friend[j].iter().filter(|&&k| !arr[k]) {
                        stack.push(k);
                    }
                }
                let cc = set.len()-1;
                for &j in set.iter() {
                    ans[j] = cc-friend[j].len()-block[j].iter().filter(|k| set.contains(k)).count();
                }
            }
        }
        ans
    };


    for i in 0..n-1 {
        print!("{} " ,ans[i])
    }
    println!("{}",ans[n-1])
}