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>

//UnionFind
//全て添字で扱うこと
struct union_find {
  std::vector<long long> parent;
  std::vector<long long> sizes;

  //コンストラクタ
  union_find(long long n) {
    for (long long i = 0; i < n; i++) {
      parent.push_back(i);
      sizes.push_back(1);
    }
  }

  //頂点nの根
  long long root_of(long long n) {
    if (parent[n] == n) return n;
    return parent[n] = root_of(parent[n]); //パス圧縮
  }

  //頂点nと間接的に繋がっている頂点の数(自身含む)
  long long size_of(long long n) {return sizes[root_of(n)];}

  //頂点a, bが間接的に繋がっているか判定
  bool is_united(long long a, long long b) {return (root_of(a) == root_of(b));}

  //頂点aを含むグラフに頂点bを含むグラフを結合
  void unite(long long a, long long b) {
    if (is_united(a, b)) return;
    if (size_of(a) < size_of(b)) std::swap(a, b); //マージテク
    sizes[root_of(a)] += size_of(b);
    parent[root_of(b)] = root_of(a);
  }
};

int main() {
  long long n, m, k;
  std::cin >> n >> m >> k;

  union_find uf(n);

  std::vector<std::vector<long long>> nx(n, std::vector<long long>(0));
  for (long long i = 0; i < m; i++) {
    long long a, b;
    std::cin >> a >> b;
    nx[a - 1].push_back(b - 1);
    nx[b - 1].push_back(a - 1);
    uf.unite(a - 1, b - 1);
  }

  std::vector<long long> bl(n, 0);
  for (long long i = 0; i < k; i++) {
    long long c, d;
    std::cin >> c >> d;
    if (uf.is_united(c - 1, d - 1)) {
      bl[c - 1]++;
      bl[d - 1]++;
    }
  }

  for (long long i = 0; i < n - 1; i++) std::cout << uf.size_of(i) - 1 - nx[i].size() - bl[i] << " ";
  std::cout << uf.size_of(n - 1) - 1 - nx[n - 1].size() - bl[n - 1] << std::endl;
}