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 mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let k: usize = s.next().unwrap().parse().unwrap();
	let mut uf = UnionFind::new(n);
	let mut edges = vec![vec![]; n];
	for _ in 0..m {
		let a = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let b = s.next().unwrap().parse::<usize>().unwrap() - 1;
		uf.unite(a, b);
		edges[a].push(b);
		edges[b].push(a);
	}
	let mut block = vec![0; n];
	for _ in 0..k {
		let a = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let b = s.next().unwrap().parse::<usize>().unwrap() - 1;
		if uf.find(a) == uf.find(b) {
			block[a] += 1;
			block[b] += 1;
		}
	}
	let mut ans = String::new();
	for i in 0..n {
		let size = uf.size(i);
		ans.push_str(&format!("{} ", size - 1 - edges[i].len() - block[i]));
	}
	ans.pop();
	println!("{}", ans);
}

pub struct UnionFind {
	data: Vec<i32>,
}

impl UnionFind {
	pub fn new(n: usize) -> UnionFind {
		UnionFind { data: vec![-1; n] }
	}

	pub fn unite(&mut self, x: usize, y: usize) -> bool {
		let x = self.find(x);
		let y = self.find(y);
		if x == y {
			false
		} else {
			let (x, y) = if self.data[x] > self.data[y] { (y, x) } else { (x, y) };
			self.data[x] += self.data[y];
			self.data[y] = x as i32;
			true
		}
	}

	pub fn find(&mut self, i: usize) -> usize {
		if self.data[i] < 0 {
			i
		} else {
			let p = self.data[i] as usize;
			self.data[i] = self.find(p) as i32;
			self.data[i] as usize
		}
	}

	pub fn size(&mut self, i: usize) -> usize {
		let i = self.find(i);
		-self.data[i] as usize
	}
}