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[100010]={0};
int d[100010]={0};
int S[100010]={0};//サイズ用配列 未実験
int Find(int a,int b){
  int r,n;
  for(r=a;p[r];r=p[r]);
  for(;n=p[a];a=n)p[a]=r;
  for(r=b;p[r];r=p[r]);
  for(;n=p[b];b=n)p[b]=r;
  return a-b?1:0;
}
void Union(int a,int b){
  if(Find(a,b)){
    for(;p[a];a=p[a]);
    for(;p[b];b=p[b]);
    d[a]<d[b]?(p[a]=b):(p[b]=a);
    if(d[a]==d[b])d[a]++;
    S[a]=S[b]=S[a]+S[b];
  }
}
int main(){
  int n,m,k,i,a,b,o,c[100010]={};
  for(i=0;i<100010;i++)S[i]=1;
  scanf("%d %d %d",&n,&m,&o);
  while(m--){
    scanf("%d %d",&a,&b);
    Union(a,b);
    c[a]++;
    c[b]++;
  }
  while(o--){
    scanf("%d %d",&a,&b);
    if(Find(a,b)==0){
      c[a]++;
      c[b]++;
    }
  }
  for(i=1;i<=n;i++){
    for(a=i;p[a];a=p[a]);
    S[i]=S[a];
  }
  ///for(i=0;i<n;i++)printf("%d %d\n",S[i+1],c[i+1]);printf("\n");
  for(i=1;i<=n;i++)printf("%d%c",S[i]-c[i]-1,n==i?'\n':' ');
  return 0;
}