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)

/**
 * @copyright (c) 2020 Daisuke Hashimoto
 */

#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using Pair = pair<int64_t, int64_t>;

// std::cout << std::setprecision(20) << 1.1 << endl;

class UnionFind {
 public:
  explicit UnionFind(const int32_t n) : parents_(n, -1) {}
  UnionFind() = delete;

  bool IsRoot(const int32_t x) const {
    return (parents_[x] < 0);
  }

  int32_t FindRoot(const int32_t x) {
    if (IsRoot(x)) {
      return x;
    }
    parents_[x] = FindRoot(parents_[x]);
    return parents_[x];
  }

  bool Unite(const int32_t x, const int32_t y) {
    int32_t root_1 = FindRoot(x);
    int32_t root_2 = FindRoot(y);
    if (root_1 == root_2) {
      return false;
    }
    if (parents_[root_1] > parents_[root_2]) {
      std::swap(root_1, root_2);
    }
    parents_[root_1] += parents_[root_2];
    parents_[root_2] = root_1;
    return true;
  }

  bool IsSame(const int32_t x, const int32_t y) {
    return FindRoot(x) == FindRoot(y);
  }

  int32_t Size(const int32_t x) {
    return -1 * parents_[FindRoot(x)];
  }

 private:
  std::vector<int32_t> parents_;
};

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int64_t N, M, K;
  cin >> N >> M >> K;
  UnionFind uf(N + 1);
  // const int64_t L = static_cast<int64_t>(T.length());
  vector<int32_t> number_of_friends(N + 1, 0);
  vector<set<int32_t>> block_list(N + 1);
  for (int64_t i = 0; i < M; ++i) {
    int32_t a, b;
    cin >> a >> b;
    uf.Unite(a, b);
    ++number_of_friends[a];
    ++number_of_friends[b];
  }
  for (int64_t i = 0; i < K; ++i) {
    int32_t c, d;
    cin >> c >> d;
    block_list[c].insert(d);
    block_list[d].insert(c);
  }
  for (int64_t i = 1; i <= N; ++i) {
    int32_t potential = uf.Size(i) - number_of_friends[i] - 1;
    for (const int32_t block : block_list[i]) {
      if (uf.IsSame(i, block)) {
        --potential;
      }
    }
    if (i != 1) {
      cout << " ";
    }
    // cerr << N << ", " << uf.Size(i) << ", " << static_cast<int32_t>(block_list[i].size()) << endl;
    cout << potential;
  }
  cout << endl;
  return 0;
}