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 (1.42.0) (judge verdict: Accepted)

use proconio::marker::Usize1;
fn main() {
    proconio::input! {
        n: usize,
        m: usize,
        k: usize,
        ab: [(Usize1,Usize1);m],
        cd: [(Usize1,Usize1);k],
    }
    let mut r=vec![0;n];
    let mut uf=vec![-1;n];
    for (a,b) in ab {
        r[a]-=1;
        r[b]-=1;
        unite(&mut uf,a,b);
    }
    for i in 0..n {
        r[i]+=count(&mut uf,i)-1;
    }
    for (c,d) in cd {
        if find(&mut uf,c)==find(&mut uf,d) {
            r[c]-=1;
            r[d]-=1;
        }
    }
    println!("{}",itertools::join(r, " "));
}
fn find(uf: &mut Vec<i32>, i: usize) -> usize {
    if uf[i]<0 {
        i
    }
    else {
        uf[i]=find(uf, uf[i]as usize)as i32;
        uf[i]as usize
    }
}
fn unite(uf: &mut Vec<i32>, i: usize, j: usize) {
    let mut i=find(uf, i);
    let mut j=find(uf, j);
    if i==j {
        return;
    }
    if uf[i]<uf[j] {
        std::mem::swap(&mut i, &mut j);
    }
    uf[i]+=uf[j];
    uf[j]=i as i32;
}
fn count(uf: &mut Vec<i32>, i: usize) -> i32 {
    let i=find(uf, i);
    -uf[i]
}