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<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<ctime>
#include<queue>
#include<set>

using namespace std;
using ll = long long;

struct UnionFind {
  vector<int> par;//parent

  vector<int>nodes;

  UnionFind(int N): par(N), nodes(N)
  {
    for(int i = 0; i < N; i++){
      par[i] = i;
      nodes[i] = 1;
    }
  }

  int root(int x) {//find the root and path compression
    if(par[x] == x) return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y){
    int rx = root(x);
    int ry = root(y);
    if(rx == ry) return;
    nodes[ry] += nodes[rx];
    par[rx] = ry;
  }

  bool same(int x, int y){
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }

};

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start--------------------
  ll n, m, k; cin >> n >> m >> k;
  vector<ll> ans(n, 0);
  UnionFind uf = (n);
  vector<ll> s(n, 0);
  for(int i = 0; i < m ; i++){
    int a, b;
    cin >> a >> b;
    a--, b--;
    if(!uf.same(a, b)) uf.unite(a, b);
    ans[a]--; ans[b]--;
  }
  for(int i = 0; i < n; i++){
    ans[i] += uf.nodes[uf.root(i)]-1;
  }
  for(int i = 0; i < k; i++){
    int a, b;
    cin >> a >> b;
    a--, b--;
    if(uf.same(a, b)) ans[a]--, ans[b]--;
  }
  for(int i = 0; i < n; i++) cout << ans[i] << ' ';
  cout << endl;
  // end-----------------------------------------------
  cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}