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, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

macro_rules! eprint {
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = write!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! eprintln {
	() => { eprintln!(""); };
	($($t:tt)*) => {{
		use ::std::io::Write;
		let _ = writeln!(::std::io::stderr(), $($t)*);
	}};
}
macro_rules! dbg {
	($v:expr) => {{
		let val = $v;
		eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
		val
	}}
}

macro_rules! mat {
	($($e:expr),*) => { Vec::from(vec![$($e),*]) };
	($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
	($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
	($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
	($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
		$a$([$i])*.$f($($t),*)
	};
	($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
		let t = $e;
		ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
	} };
}

pub fn readln() -> String {
	let mut line = String::new();
	::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
	line
}

macro_rules! read {
	($($t:tt),*; $n:expr) => {{
		let stdin = ::std::io::stdin();
		let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
			let line = line.unwrap();
			let mut it = line.split_whitespace();
			_read!(it; $($t),*)
		}).collect::<Vec<_>>();
		ret
	}};
	($($t:tt),*) => {{
		let line = readln();
		let mut it = line.split_whitespace();
		_read!(it; $($t),*)
	}};
}

macro_rules! _read {
	($it:ident; [char]) => {
		_read!($it; String).chars().collect::<Vec<_>>()
	};
	($it:ident; [u8]) => {
		Vec::from(_read!($it; String).into_bytes())
	};
	($it:ident; usize1) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
	};
	($it:ident; [usize1]) => {
		$it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
	};
	($it:ident; [$t:ty]) => {
		$it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
	};
	($it:ident; $t:ty) => {
		$it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
	};
	($it:ident; $($t:tt),+) => {
		($(_read!($it; $t)),*)
	};
}

pub fn main() {
	let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: usize = 1_000_000_007;
const INF: i64 = std::i64::MAX/2;

pub trait SetMinMax {
	fn setmin(&mut self, v: Self) -> bool;
	fn setmax(&mut self, v: Self) -> bool;
}
impl<T> SetMinMax for T where T: PartialOrd {
	fn setmin(&mut self, v: T) -> bool {
		*self > v && { *self = v; true }
	}
	fn setmax(&mut self, v: T) -> bool {
		*self < v && { *self = v; true }
	}
}

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

impl UnionFind {
	pub fn new(n: usize) -> Self{
		UnionFind {
			data: vec![-1;n]
		}
	}
	pub fn root(&mut self, x: usize) -> usize{
		if self.data[x]<0 {
			x
		}
		else {
			let p = self.data[x] as usize;
			self.data[x] = self.root(p) as i32;
			self.data[x] as usize
		}
	}
	pub fn unite(&mut self, x: usize, y: usize){
		let rx = self.root(x);
		let ry = self.root(y);
		if rx!=ry {
			let (rx,ry) = if self.data[rx]>self.data[ry] {
				(rx,ry)
			}
			else{
				(ry,rx)
			};
			self.data[rx] += self.data[ry];
			self.data[ry] = rx as i32;
		}
	}
	pub fn same(&mut self, x: usize, y: usize) ->  bool{
		self.root(x) == self.root(y)
	}
	pub fn size(&mut self, x: usize) -> i32 {
		let r = self.root(x);
		- self.data[r]
	}
}

fn solve() {
	let (n,m,k) = read!(usize,usize,usize);
	let ab = read!([usize];m);
	let cd = read!([usize];k);
	let mut uf = UnionFind::new(n);
	let mut num = vec![1;n];
	for i in 0..m {
		let s = ab[i][0]-1;
		let t = ab[i][1]-1;
		uf.unite(s,t);
		num[s]+=1;
		num[t]+=1;
	}
	let mut v = vec![vec![];n];
	for i in 0..k {
		let s = cd[i][0]-1;
		let t = cd[i][1]-1;
		v[s].push(t);
		v[t].push(s);
	}
	for i in 0..n{
		let mut ans = uf.size(i)-num[i];
		for &j in &v[i]{
			if uf.same(i,j) {
				ans -= 1;
			}
		}
		print!("{} ", ans);
	}
}

fn run() {
    solve();
}