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)

fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let n: usize = sc.read();
    let m: usize = sc.read();
    let k: usize = sc.read();
    let mut uf = UnionFind::new(n);
    let mut graph = vec![vec![]; n];
    for _ in 0..m {
        let a = sc.read::<usize>() - 1;
        let b = sc.read::<usize>() - 1;
        graph[a].push(b);
        graph[b].push(a);
        uf.unite(a, b);
    }
    let mut block_graph = vec![vec![]; n];
    for _ in 0..k {
        let a = sc.read::<usize>() - 1;
        let b = sc.read::<usize>() - 1;
        block_graph[a].push(b);
        block_graph[b].push(a);
    }

    for i in 0..n {
        let root = uf.find(i);
        let friends = graph[i].len();
        let candidates = uf.sizes[root];
        let mut blocked = 0;
        for &next in block_graph[i].iter() {
            if uf.find(next) == root {
                blocked += 1;
            }
        }
        let ans = candidates - friends - blocked - 1;

        if i > 0 {
            sc.write(" ");
        }
        sc.write(ans.to_string());
    }
    sc.write("\n");
}
pub struct UnionFind {
    parent: Vec<usize>,
    sizes: Vec<usize>,
    size: usize,
}

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        UnionFind {
            parent: (0..n).map(|i| i).collect::<Vec<usize>>(),
            sizes: vec![1; n],
            size: n,
        }
    }

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

    pub fn unite(&mut self, x: usize, y: usize) -> bool {
        let parent_x = self.find(x);
        let parent_y = self.find(y);
        if parent_x == parent_y {
            return false;
        }

        let (large, small) = if self.sizes[parent_x] < self.sizes[parent_y] {
            (parent_y, parent_x)
        } else {
            (parent_x, parent_y)
        };

        self.parent[small] = large;
        self.sizes[large] += self.sizes[small];
        self.sizes[small] = 0;
        self.size -= 1;
        return true;
    }
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> IO<R, W> {
        IO(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: std::ops::Deref<Target = str>>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}