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 <malloc.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

//ABC157D
int root(int *pare, int x) {
  if(pare[x-1]==x) return x;
  return root(pare, pare[x-1]);
}

void connect(int *pare, int *cou, int a, int b) {
  int tmp;
  if(pare[a-1]==a) tmp=a, a=b, b=tmp;
  int ra=root(pare, a), rb=root(pare, b);
  if(ra!=rb) cou[ra-1]+=cou[rb-1];
  pare[a-1]=ra, pare[rb-1]=ra;
}

int main(void) {
  int n, m, k, i;
  scanf("%d %d %d", &n, &m, &k);
  int a[m], b[m], c, d, pare[n], friend[n], block[n], cou[n];
  for(i=0; i<n; i++)
    pare[i]=i+1, friend[i]=0, block[i]=0, cou[i]=1;

  for(i=0; i<m; i++){
    scanf(" %d %d", &a[i], &b[i]);
    connect(pare, cou, a[i], b[i]);
    friend[a[i]-1]++, friend[b[i]-1]++;
  }
  for(i=0; i<k; i++){
    scanf(" %d %d", &c, &d);
    if(root(pare, c)==root(pare, d)) block[c-1]++, block[d-1]++;
  }

  for(i=1; i<=n; i++)
    printf("%d ", cou[root(pare, i)-1]-block[i-1]-friend[i-1]-1);
}