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)

use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: ident) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

struct UnionFind {
    parents: Vec<usize>,
}
impl UnionFind {
    pub fn new(n: usize) -> Self {
        let par = (0..n).collect();
        UnionFind { parents: par }
    }
    pub fn root(&mut self, x: usize) -> usize {
        if self.parents[x] == x {
            return x;
        }
        let par = self.parents[x];
        self.parents[x] = self.root(par);
        self.parents[x]
    }
    pub fn union(&mut self, x: usize, y: usize) {
        let rx = self.root(x);
        let ry = self.root(y);
        if rx == ry {
            return;
        }
        self.parents[rx] = ry;
    }
    pub fn is_same(&mut self, x: usize, y: usize) -> bool {
        let rx = self.root(x);
        let ry = self.root(y);
        rx == ry
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    let mut it = s.split_whitespace();
    let (n, m, k) = parse!(it, 3);
    let mut res = vec![0; n];
    let mut ren = vec![0; n];
    let mut uf = UnionFind::new(n);
    for i in 0..m {
        let (x, y): (usize, usize) = parse!(it, 2);
        uf.union(x - 1, y - 1);
        ren[x - 1] += 1;
        ren[y - 1] += 1;
    }

    for _ in 0..k {
        let (x, y): (usize, usize) = parse!(it, 2);
        let x = x - 1;
        let y = y - 1;
        if uf.is_same(x, y) {
            res[x] += 1;
            res[y] += 1;
        }
    }
    for i in 0..n {
        uf.root(i);
    }
    let mut counts = std::collections::HashMap::new();
    for i in 0..n {
        let r = uf.root(i);
        let mut c = 1;
        if let Some(t) = counts.get(&r) {
            c = *t + 1;
        }
        counts.insert(r, c);
    }
    let ret = (0..n)
        .map(|x| {
            let a = *counts.get(&uf.root(x)).unwrap();
            let b = res[x];
            a - b - ren[x] - 1
        })
        .map(|x: usize| x.to_string())
        .collect::<Vec<_>>()
        .join(" ");
    println!("{}", ret);
}