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>
#define int long long
using namespace std;

class UnionFind {
private:
  vector<int> data;

  bool isSame(int x, int y) { return find(x) == find(y); }

  bool doUnite(int x, int y) {
    x = find(x), y = find(y);
    if (x == y)
      return (false);
    if (data[x] > data[y])
      swap(x, y);
    data[x] += data[y];
    data[y] = x;
    return (true);
  }

  int getFind(int k) {
    if (data[k] < 0)
      return (k);
    return (data[k] = find(data[k]));
  }

  int getSize(int k) { return (-data[find(k)]); }

public:
  UnionFind(int N) { data.resize(N, -1); }

  int same(int x, int y) { return isSame(x, y); }
  bool unite(int x, int y) { return doUnite(x, y); }
  int find(int k) { return getFind(k); }
  int size(int k) { return getSize(k); }
};

signed main() {
  int N, M, K;
  cin >> N >> M >> K;

  vector<set<int>> dame(N);
  UnionFind uf(N);

  for (int i = 0; i < M; i++) {
    int A, B;
    cin >> A >> B;
    A--, B--;
    dame[A].insert(B);
    dame[B].insert(A);
    uf.unite(A, B);
  }

  for (int i = 0; i < K; i++) {
    int C, D;
    cin >> C >> D;
    C--, D--;
    if (!uf.same(C, D)) {
      continue;
    }
    dame[C].insert(D);
    dame[D].insert(C);
  }

  for (int i = 0; i < N; i++) {
    if (i != 0) {
      cout << " ";
    }
    cout << uf.size(i) - dame[i].size() - 1;
  }

  cout << endl;
  return 0;
}