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::*;

mod disjoint_set {
    pub struct DisjointSet {
        parent: Vec<usize>,
        size: Vec<usize>,
    }
    impl DisjointSet {
        pub fn new(n: usize) -> DisjointSet {
            DisjointSet {
                parent: (0..n).collect(),
                size: vec![1; n],
            }
        }
        fn root(&mut self, x: usize) -> usize {
            if x == self.parent[x] {
                x
            } else {
                let tmp = self.parent[x];
                self.parent[x] = self.root(tmp);
                self.parent[x]
            }
        }
        pub fn merge(&mut self, x: usize, y: usize) {
            let x = self.root(x);
            let y = self.root(y);
            if x == y {
                return;
            }
            if self.size[x] > self.size[y] {
                self.size[x] += self.size[y];
                self.parent[y] = x;
            } else {
                self.size[y] += self.size[x];
                self.parent[x] = y;
            }
        }
        pub fn same(&mut self, x: usize, y: usize) -> bool {
            self.root(x) == self.root(y)
        }
        pub fn size(&mut self, x: usize) -> usize {
            let x = self.root(x);
            self.size[x]
        }
    }
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: usize = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();

    // 条件4: 友人関係のグラフ上でa, bは連結成分
    let mut ds = disjoint_set::DisjointSet::new(100010);
    let mut graph = vec![Vec::new(); 100010];
    for _ in 0..m {
        let a = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        let b = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        ds.merge(a, b);
        graph[a].push(b);
        graph[b].push(a);
    }
    for _ in 0..k {
        let c: usize = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        let d: usize = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        graph[c].push(d);
        graph[d].push(c);
    }

    let mut out = Vec::new();
    for i in 0..n {
        let mut ans = ds.size(i);
        for &v in graph[i].iter() {
            if ds.same(i, v) {
                ans -= 1;
            }
        }
        writeln!(out, "{}", ans - 1).ok();
    }
    stdout().write_all(&out).ok();
}