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: Time Limit Exceeded)

macro_rules! scan_sub {
    ( $itr:ident, $t:ty ) => {
        let mut buf = String::new();
        let $itr = {
            while buf.trim().len() == 0 {
                buf = String::new();
                std::io::stdin().read_line(&mut buf).unwrap();
            }
            buf.split_whitespace().map(|n| n.parse::<$t>().unwrap())
        };
    };
}

macro_rules! scan {
    ( $x:ident; Vec<$t:ty> ) => {
        scan_sub!(itr, $t);
        let $x = itr.collect::<Vec<_>>();
    };
    ( $( $x:ident ),*; $t:ty ) => {
        scan_sub!(itr, $t);
        let mut itr = itr;
        $(
            let $x = itr.next().unwrap();
        )*
    };
}

fn find(x: usize, par: &mut Vec<usize>) -> usize {
    if par[x] == x {
        return x;
    }

    par[x] = find(par[x], par);
    par[x]
}

fn unite(x: usize, y: usize, par: &mut Vec<usize>, rank: &mut Vec<usize>) {
    let x = find(x, par);
    let y = find(y, par);
    if x == y {return;}

    if rank[x] < rank[y] {
        par[x] = y;
    } else {
        par[y] = x;
        if rank[x] == rank[y] {
            rank[x] += 1;
        }
    }
}

fn same(x: usize, y: usize, par: &mut Vec<usize>) -> bool {
    find(x, par) == find(y, par)
}

fn main() {
    scan!(n, m, k; usize);
    let mut friend_list = (0..n).map(|_| Vec::with_capacity(n)).collect::<Vec<_>>();
    let mut block_list = (0..n).map(|_| Vec::with_capacity(n)).collect::<Vec<_>>();
    let mut par = (0..n).collect::<Vec<_>>();
    let mut rank = vec![0; n];

    for _ in 0..m {
        scan!(a, b; usize);
        let (a, b) = (a-1, b-1);
        friend_list[a].push(b);
        friend_list[b].push(a);
        unite(a, b, &mut par, &mut rank);
    }

    for _ in 0..k {
        scan!(c, d; usize);
        let (c, d) = (c-1, d-1);
        block_list[c].push(d);
        block_list[d].push(c);
    }

    let mut par_size = vec![0; n];
    for i in 0..n {
        par_size[find(i, &mut par)] += 1;
    }

    for i in 0..n {
        let mut t = par_size[par[i]];
        friend_list[i].iter().map(|&j| {
            if same(i, j, &mut par) {
                t -= 1;
            }
        }).collect::<Vec<_>>();
        block_list[i].iter().map(|&j| {
            if same(i, j, &mut par) {
                t -= 1;
            }
        }).collect::<Vec<_>>();
        print!("{} ", t-1);
    }
    println!("");
}