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::collections::HashMap;

fn read1<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn readn<T: std::str::FromStr>(delimiter: &str) -> Vec<T> {
    let s = read1::<String>();
    s.split(delimiter).map(|e| e.parse().ok().unwrap()).collect::<Vec<T>>()
}

pub struct UnionFind {
    pub par: Vec<usize>,
    pub rank: Vec<usize>,
}

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        let par = (0..n).collect::<Vec<usize>>();
        let rank = vec![0; n];
        UnionFind { par: par, rank: rank }
    }

    pub fn root(&mut self, x: usize) -> usize {
        let cand = self.par[x];
        if cand == x {
            return x;
        } else {
            let ret = self.root(cand);
            self.par[x] = ret;
            return ret;
        }
    }

    pub fn same(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }

    pub fn unite(&mut self, x: usize, y: usize) {
        let root_x = self.root(x);
        let root_y = self.root(y);
        if root_x == root_y { return; }

        if self.rank[root_x] < self.rank[root_y] {
            self.par[root_x] = self.par[root_y];
        } else {
            self.par[root_y] = self.par[root_x];
            if self.rank[root_y] == self.rank[root_x] {
                self.rank[root_x] += 1;
            }
        }
    }
}

fn main() {
    let nmk = readn::<usize>(" ");
    let n = nmk[0];
    let m = nmk[1];
    let k = nmk[2];

    let mut uf = UnionFind::new(n);
    let mut loss = vec![1; n];
    for _ in 0..m {
        let ab = readn::<usize>(" ");
        let a = ab[0] - 1;
        let b = ab[1] - 1;
        uf.unite(a, b);
        loss[a] += 1;
        loss[b] += 1;
    }

    let mut dict = HashMap::new();
    for i in 0..n {
        let count = dict.entry(uf.root(i)).or_insert(0usize);
        *count += 1;
    }

    for _ in 0..k {
        let cd = readn::<usize>(" ");
        let c = cd[0] - 1;
        let d = cd[1] - 1;
        if uf.same(c, d) {
            loss[c] += 1;
            loss[d] += 1;
        }
    }

    let mut ans = "".to_string();
    for i in 0..n {
        let num = dict.get(&uf.root(i)).unwrap() - loss[i];
        ans = ans + &(num.to_string()) + " ";
    }
    println!("{}", ans);
}