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 (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

pub struct UnionFind {
    n: usize,
    root: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    pub fn new(n: usize) -> Self {
        let root = (0..n).collect();
        let rank = vec![0; n];
        let size = vec![1; n];
        Self {
            n,
            root,
            rank,
            size,
        }
    }

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

    pub fn unite(&mut self, x: usize, y: usize) {
        if x >= self.n || y >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        let y_root = self.find(y);
        if x_root == y_root {
            return;
        }
        if self.rank[x_root] < self.rank[y_root] {
            self.root[x_root] = y_root;
            self.size[y_root] += self.size[x_root];
        } else {
            self.root[y_root] = x_root;
            self.size[x_root] += self.size[y_root];
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }

    pub fn size(&mut self, x: usize) -> usize {
        if x >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        self.size[x_root]
    }

    pub fn is_same(&mut self, x: usize, y: usize) -> bool {
        if x >= self.n || y >= self.n {
            panic!();
        }
        self.find(x) == self.find(y)
    }
}

fn solve() {
    input! {
        n: usize, m: usize, k: usize,
        abv: [(Usize1, Usize1); m],
        cdv: [(Usize1, Usize1); k]
    };

    let mut friend_relations = UnionFind::new(n);
    let mut friends = vec![0; n];
    for (a, b) in abv {
        friend_relations.unite(a, b);
        friends[a] += 1;
        friends[b] += 1;
    }
    let mut blocks = vec![Vec::new(); n];
    for (c, d) in cdv {
        blocks[c].push(d);
        blocks[d].push(c);
    }
    let res = (0..n)
        .map(|u| {
            let all = friend_relations.size(u) - 1;
            let mut res = all - friends[u];
            for &b in &blocks[u] {
                if friend_relations.is_same(u, b) {
                    res -= 1;
                }
            }
            res
        })
        .join(" ");
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}