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)

use std::collections::{
    HashMap, HashSet
};

#[derive(Debug)]
struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        let mut p = vec![0; n];
        for i in 0..n {
            p[i] = i;
        }
        UnionFind {
            parent: p,
            rank: vec![0; n],
            size: vec![1; n],
        }
    }

    fn root(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let p = self.parent[x];
            self.parent[x] = self.root(p);
            self.parent[x]
        }
    }

    fn unite(&mut self, a: usize, b:usize) {
        let ra = self.root(a);
        let rb = self.root(b);
        if self.rank[ra] > self.rank[rb] {
            self.parent[rb] = ra;
            self.size[ra] += self.size[rb];
        } else if self.rank[ra] < self.rank[rb] {
            self.parent[ra] = rb;
            self.size[rb] += self.size[ra];
        } else if ra != rb {
            self.parent[rb] = ra;
            self.size[ra] += self.size[rb];
            self.rank[ra] += 1;
        }
    }

    fn size (&mut self, x: usize) -> usize {
        let rx = self.root(x);
        self.size[rx]
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.root(a) == self.root(b)
    }

}


fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|i| i.parse().ok().unwrap())
        .collect()
}

fn main() {
    let input = read_vec::<usize>();
    let n = input[0];
    let m = input[1];
    let k = input[2];

    let mut uf = UnionFind::new(n);
    let mut friends: HashMap<usize, HashSet<usize>> = HashMap::new();
    let mut blocks: HashMap<usize, HashSet<usize>> = HashMap::new();

    for i in 0..n {
        friends.insert(i, HashSet::new());
        blocks.insert(i, HashSet::new());
    }

    for _ in 0..m {
        let input = read_vec::<usize>();
        uf.unite(input[0]-1, input[1]-1);
        friends.get_mut(&(input[0]-1)).unwrap().insert(input[1]-1);
        friends.get_mut(&(input[1]-1)).unwrap().insert(input[0]-1);
        
    }

    for _ in 0..k {
        let input = read_vec::<usize>();
        blocks.get_mut(&(input[0]-1)).unwrap().insert(input[1]-1);
        blocks.get_mut(&(input[1]-1)).unwrap().insert(input[0]-1);
        

    }

    let mut ans: Vec<usize> = vec![];

    for i in 0..n {
        let s = uf.size(i);
        let f = &friends[&i];
        let b = &blocks[&i];
        let mut c = 0;
        for j in b {
            if uf.same(i, *j) {
                c += 1;
            }
        }
        ans.push(s-f.len()-c-1);
    }

    let mut s = String::new();
    for a in ans {
        s = format!("{} {}",&s, a);
        // s += " ";
    }
    println!("{}", s.trim());
}