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(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_macros)]
#![allow(unused_imports)]

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

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    }
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

macro_rules! chmax {
    ($x:expr, $e:expr) => {
        if $x < $e {
            $x = $e;
        }
    };
}
macro_rules! chmin {
    ($x:expr, $e:expr) => {
        if $x > $e {
            $x = $e;
        }
    };
}

const INF: i64 = 1 << 60;

struct Solver {
    users: Vec<i64>,
}

impl Solver {
    fn new(N: usize) -> Solver {
        Solver { users: vec![-1; N] }
    }
    fn dfs(&mut self, i: usize, l: i64, g: &Vec<Vec<usize>>) {
        if self.users[i] != -1 {
            return;
        }
        self.users[i] = l;
        for &i in &g[i] {
            self.dfs(i, l, g);
        }
    }
}

use std::iter::IntoIterator;

fn join_string<T, S>(elems: T) -> String
where
    T: IntoIterator<Item = S>,
    S: ToString,
{
    elems
        .into_iter()
        .map(|i| i.to_string())
        .collect::<Vec<String>>()
        .join(" ")
}

fn main() {
    input!{
        N: usize,
        M: usize,
        K: usize,
        AB: [(usize1, usize1); M],
        CD: [(usize1, usize1); K],
    }
    let mut to = vec![Vec::new(); N];
    for &(A, B) in &AB {
        to[A].push(B);
        to[B].push(A);
    }
    let mut s = Solver::new(N);
    let mut l = 0;
    for i in 0..N {
        if s.users[i] != -1 {
            continue;
        }
        s.dfs(i, l, &to);
        l += 1;
    }
    let mut counter = vec![0; l as usize];
    for i in 0..N {
        counter[s.users[i] as usize] += 1;
    }
    let mut ans = vec![0; N];
    for i in 0..N {
        ans[i] += counter[s.users[i] as usize] - 1;
    }
    for i in 0..M {
        let &(A, B) = &AB[i];
        if s.users[A] != s.users[B] {
            continue;
        }
        ans[A] -= 1;
        ans[B] -= 1;
    }
    for i in 0..K {
        let &(C, D) = &CD[i];
        if s.users[C] != s.users[D] {
            continue;
        }
        ans[C] -= 1;
        ans[D] -= 1;
    }
    println!("{}", join_string(ans));
}