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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (int)n; i++)
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long int ll;
typedef vector<int > vecint;
typedef vector<bool > vecbool;
typedef vector<string > vecstr;
int n, m, k;
vector<unordered_set<int>> fr;
vector<unordered_set<int>> nonfr;
vector<unordered_set<int>> tree;
vecbool seen;
vecint tmap;
void dfs(int i, int x){
  //loop for friend x
  if(!seen[x]){
    seen[x] = true;
    // tree[i].insert(x);
    tmap[x] = i;
  }
  for(auto itr = fr[x].begin(); itr!= fr[x].end();itr++){
    if(seen[*itr])continue;
    seen[*itr] = true;
    // tree[i].insert(*itr);
    tmap[*itr] = i;
    dfs(i, *itr);
  }
}
int main()
{
  cin >> n >> m >> k;
  ///////
  fr.resize(n + 1);
  nonfr.resize(n + 1);
  int a, b;
  REP(i,m){
    cin >> a >> b;
    fr[a].insert(b);
    fr[b].insert(a);
    nonfr[a].insert(b);
    nonfr[b].insert(a);
  }
  REP(i,k){
    cin >> a >> b;
    nonfr[a].insert(b);
    nonfr[b].insert(a);
  }
  ///////
  seen.resize(n + 1);
  fill(all(seen), false);
  seen[0] = true;
  tmap.resize(n + 1);
  fill(all(tmap), -1);
  ///////
  int ini = 1;
  int nt = 0;
  bool ok = false;
  while(!ok){
    // tree.resize(ini);
    // tree.push_back(unordered_set<int >());
    dfs(nt,ini);
    nt++;
    ok = true;
    REP(i,n+1){
      if(!seen[i]){
	ini = i;
	ok = false;
	break;
      }
    }
  }
  // for(int i = 0; i < tree.size(); i ++){
  //   cout << "tree "<<i << endl;
  //   for(auto itr = tree[i].begin(); itr != tree[i].end();itr++){
  //     cout << *itr << " ";
  //   }
  //   cout << endl;
  // }
  // for(int i = 1; i < nonfr.size(); i ++){
  //   cout << "nonfr "<<i << endl;
  //   for(auto itr = nonfr[i].begin(); itr != nonfr[i].end();itr++){
  //     cout << *itr << " ";
  //   }
  //   cout << endl;
  // }
  vecint num(nt + 1,0);
  for(int i = 1; i <= n; i++){
    num[tmap[i]]++;
  }
  for(int i = 1; i <= n; i ++){
    // cout << tmap[i] << endl;
    int ans = 0;
    // ans = tree[tmap[i]].size() - nonfr[i].size() - 1;
    ans = num[tmap[i]] - nonfr[i].size() - 1;
    cout << ans <<" ";
  }
  cout << endl;
}