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)

macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; ( iter = $ iter : ident , $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut $ iter = s . split_whitespace ( ) ; input_inner ! { $ iter , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut iter = s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; }
macro_rules ! input_inner { ( $ iter : expr ) => { } ; ( $ iter : expr , ) => { } ; ( $ iter : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , mut $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , mut $ var : usize $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident $ ( $ r : tt ) * ) => { input_inner ! { $ iter , $ var : usize $ ( $ r ) * } } ; }
macro_rules ! read_value { ( $ iter : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ iter , $ t ) ) ,* ) } ; ( $ iter : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ iter , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ iter : expr , { chars : $ base : expr } ) => { read_value ! ( $ iter , String ) . chars ( ) . map ( | c | ( c as u8 - $ base as u8 ) as usize ) . collect ::< Vec < usize >> ( ) } ; ( $ iter : expr , { char : $ base : expr } ) => { read_value ! ( $ iter , { chars : $ base } ) [ 0 ] } ; ( $ iter : expr , chars ) => { read_value ! ( $ iter , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ iter : expr , char ) => { read_value ! ( $ iter , chars ) [ 0 ] } ; ( $ iter : expr , usize1 ) => { read_value ! ( $ iter , usize ) - 1 } ; ( $ iter : expr , $ t : ty ) => { $ iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) } ; }
pub mod at {
    pub struct UnionFind {
        parents: Vec<isize>,
    }
    use std::collections::HashMap;
    impl UnionFind {
        pub fn new(n: usize) -> UnionFind {
            let parents = vec![-1; n];
            UnionFind { parents: parents }
        }
        pub fn find(&mut self, x: usize) -> usize {
            if self.parents[x] < 0 {
                x
            } else {
                let xx = self.parents[x] as usize;
                let y = self.find(xx);
                self.parents[x] = y as isize;
                y
            }
        }
        pub fn unite(&mut self, x: usize, y: usize) -> bool {
            use std::mem::swap;
            let mut x = self.find(x);
            let mut y = self.find(y);
            if x == y {
                return false;
            }
            if self.parents[x] > self.parents[y] {
                swap(&mut x, &mut y);
            }
            self.parents[x] += self.parents[y];
            self.parents[y] = x as isize;
            true
        }
        pub fn size(&mut self, x: usize) -> usize {
            let x = self.find(x);
            (-self.parents[x]) as usize
        }
        pub fn same(&mut self, x: usize, y: usize) -> bool {
            self.find(x) == self.find(y)
        }
        pub fn members(&mut self, x: usize) -> Vec<usize> {
            let root = self.find(x);
            (0..self.parents.len())
                .into_iter()
                .filter(|i| self.find(*i) == root)
                .collect::<Vec<usize>>()
        }
        pub fn roots(&mut self) -> Vec<usize> {
            (0..self.parents.len())
                .into_iter()
                .filter(|i| self.parents[*i] < 0)
                .collect::<Vec<usize>>()
        }
        pub fn all_group_members(&mut self) -> HashMap<usize, Vec<usize>> {
            let mut groups_map = HashMap::new();
            for x in 0..self.parents.len() {
                let r = self.find(x);
                groups_map.entry(r).or_insert(vec![]).push(x);
            }
            groups_map
        }
    }
}
use at::*;
fn main() {
    input! { n, m, k, ab:[(usize1,usize1); m], cd:[(usize1,usize1); k] };
    let mut uf = UnionFind::new(n);
    let mut ans = vec![0; n];
    for &(a, b) in &ab {
        uf.unite(a, b);
    }
    for i in 0..n {
        ans[i] = uf.size(i) - 1;
    }
    for &(a, b) in &ab {
        ans[a] -= 1;
        ans[b] -= 1;
    }
    for &(c, d) in &cd {
        if uf.same(c, d) {
            ans[c] -= 1;
            ans[d] -= 1;
        }
    }
    for i in 0..n {
        print!("{} ", ans[i]);
    }
    println!("");
}