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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;
using Graph = vector<vector<int>>;

int main() {
  int N, M, K;
  cin >> N >> M >> K;
  vector<bool> seen(N, false);
  vector<int> gp(N);
  Graph g(N);
  Graph gg(N);
  for(int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    g.at(x-1).push_back(y-1);
    g.at(y-1).push_back(x-1);
  }
  for(int i = 0; i < K; i++) {
    int x, y;
    cin >> x >> y;
    gg.at(x-1).push_back(y-1);
    gg.at(y-1).push_back(x-1);
  }
  stack<int> todo;
  todo.push(0);
  seen.at(0)=true;
  int t=0;
  int v=0;
  gp.at(0) = 0;
  vector<int> gpnum(N);
  gpnum.at(0)++;
  while(todo.size() != 0) {
    v = todo.top();
    todo.pop();
    for(int i:g.at(v)){
      if(!seen.at(i)){
        seen.at(i) = true;
        gp.at(i) = t;
        gpnum.at(t)++;
        todo.push(i);
      }
    }
    if(todo.size() == 0) {
      t++;
      for(int i = 0; i < N; i++) {
        if(!seen.at(i)){
          seen.at(i) = true;
          gp.at(i) = t;
          gpnum.at(t)++;
          todo.push(i);
          break;
        }
      }
    }
  }
  for(int i = 0; i < N; i++) {
    int cnt = 0;
    for(int j:gg.at(i)) {
      if(gp.at(j) == gp.at(i)) cnt++;
    }
    cout << gpnum.at(gp.at(i)) - g.at(i).size() -1 -cnt << " ";
  }
}