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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<usize>();
    let (n, m, k) = (v[0], v[1], v[2]);
    let mut uft = UnionFindTree::new(n);
    let mut count = vec![0; n];

    for i in 0..m {
        let v = read_vec::<usize>();
        let (a, b) = (v[0] - 1, v[1] - 1);
        uft.unite(a, b);
        count[a] += 1;
        count[b] += 1;
    }
    let mut blocks = vec![];
    for i in 0..k {
        let v = read_vec::<usize>();
        let (a, b) = (v[0] - 1, v[1] - 1);
        blocks.push((a, b));
    }

    let mut ans = vec![0i64; n];
    for i in 0..n {
        ans[i] += (uft.get_size(i) - count[i] - 1) as i64;
    }

    for (a, b) in blocks {
        if uft.same(a, b) {
            ans[a] -= 1;
            ans[b] -= 1;
        }
    }
    for num in ans {
        print!("{} ", num);
    }
    println!("");
}

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(|e| e.parse().ok().unwrap())
        .collect()
}

#[derive(Debug, Clone)]
struct UnionFindTree {
    parent: Vec<isize>,
    size: Vec<usize>,
    height: Vec<u64>,
}

impl UnionFindTree {
    fn new(n: usize) -> UnionFindTree {
        UnionFindTree {
            parent: vec![-1; n],
            size: vec![1usize; n],
            height: vec![0u64; n],
        }
    }

    fn find(&mut self, index: usize) -> usize {
        if self.parent[index] == -1 {
            return index;
        }
        let idx = self.parent[index] as usize;
        let ret = self.find(idx);
        self.parent[index] = ret as isize;
        ret
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.find(x) == self.find(y)
    }

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

    fn unite(&mut self, index0: usize, index1: usize) -> bool {
        let a = self.find(index0);
        let b = self.find(index1);
        if a == b {
            false
        } else {
            if self.height[a] > self.height[b] {
                self.parent[b] = a as isize;
                self.size[a] += self.size[b];
            } else if self.height[a] < self.height[b] {
                self.parent[a] = b as isize;
                self.size[b] += self.size[a];
            } else {
                self.parent[b] = a as isize;
                self.size[a] += self.size[b];
                self.height[a] += 1;
            }
            true
        }
    }
}