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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std;

struct UnionFind {
  vector<int> data;
  UnionFind(int size) : data(size, -1) { }
  bool unionSet(int x, int y) {
    x = root(x); y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y]; data[y] = x;
    }
    return x != y;
  }
  bool findSet(int x, int y) {
    return root(x) == root(y);
  }
  int root(int x) {
    return data[x] < 0 ? x : data[x] = root(data[x]);
  }
  int size(int x) {
    return -data[root(x)];
  }
};

int main(){
  int N,M,K;
  cin >> N >> M >> K;
  UnionFind tree(N);
  vector<vector<int>> follow(N);
  vector<vector<int>> block(N);
  
  for(int i = 0; i < M; i++){
    int tmp1,tmp2;
    cin >> tmp1 >> tmp2;
    tmp1--; tmp2--;
    follow.at(tmp1).push_back(tmp2);
    follow.at(tmp2).push_back(tmp1);
    tree.unionSet(tmp1,tmp2);
  }
  for(int i = 0; i < K; i++){
    int tmp1,tmp2;
    cin >> tmp1 >> tmp2;
    tmp1--; tmp2--;
    block.at(tmp1).push_back(tmp2);
    block.at(tmp2).push_back(tmp1);
  }
  
  for(int i = 0; i < N; i++){
    // 同じ集合
    // フォロー数
    // 同じ集合かつブロックしている人をひく
    int ans = tree.size(i);
    ans -= follow.at(i).size();
    for(int j: block.at(i)){
      if(tree.root(i) == tree.root(j)){
        ans--;
      }
    }
    // 自分を引く
    ans--;
    if(i < N-1) cout << ans << " ";
    else cout << ans << endl;
  }
  
}