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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int p[100005],count[100005];
int root(int n){
    if(p[n]!=n)
        p[n]=root(p[n]);
    return p[n];
}
void uni(int x,int y){
    x=root(x);y=root(y);
    if(x!=y){
        p[x]=y;
        count[y]+=count[x];
    }
    return ;
}
int main(){
    int n,m,k,i;
    scanf("%d%d%d",&n,&m,&k);
    int a[100005],b[100005],c[100005],d[100005],e[100005];
    for(i=0;i<m;i++){
        scanf("%d%d",&a[i],&b[i]);
        a[i]--;b[i]--;
    }
    for(i=0;i<k;i++){
        scanf("%d%d",&c[i],&d[i]);
        c[i]--;
        d[i]--;
    }
    for(i=0;i<n;i++){
        p[i]=i;
        count[i]=1;
    }
    for(i=0;i<m;i++)
        uni(a[i],b[i]);
    for(i=0;i<n;i++) e[i]=1;
    for(i=0;i<m;i++){
        if(root(a[i])==root(b[i])){
            e[a[i]]++;e[b[i]]++;
        }
    }
    for(i=0;i<k;i++){
        if(root(c[i])==root(d[i])){
            e[c[i]]++;e[d[i]]++;
        }
    }
    for(i=0;i<n;i++)
        printf("%d ",count[root(i)]-e[i]);
    printf("\n");
    return 0;
}