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++ (GCC 9.2.1) (judge verdict: Accepted)

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

// ----------- define --------------
#define int long long
#define vi vector<int>
#define ii pair<int,int>
#define fi first
#define sc second
#define stoi stoll
#define popcnt __builtin_popcount
#define getbit(x, k) ((x >> k) & 1)
#define all(x) (x).begin(),(x).end()
// ---------------------------------

const int N = 100005;

int root[N], hang[N];

int find(int i) {
  if (i == root[i]) return i;
  return root[i] = find(root[i]);
}

int join(int u, int v) {
  u = find(u); v = find(v);
  if (u == v) return false;
  if (hang[u] < hang[v]) swap(u, v);
  hang[u] += hang[v];
  root[v] = u;
  return true;
}

void Main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    root[i] = i;
    hang[i] = 1;
  }
  vector<int> fr(n + 1);
  vector<vi> bl(n + 1);
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    join(u, v);
    fr[u]++;
    fr[v]++;
  }
  for (int i = 1; i <= k; i++) {
    int u, v;
    cin >> u >> v;
    bl[u].push_back(v);
    bl[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    int ans = hang[find(i)] - 1;
    ans -= fr[i];
    for (auto &j : bl[i]) {
      if (find(i) == find(j)) ans--;
    }
    cout << ans << ' ';
  }
  cout << '\n';
}

signed main() {
  // freopen("in" , "r", stdin );
  // freopen("out", "w", stdout);
  ios::sync_with_stdio(0); cin.tie(0);
  int T = 1;
  // cin >> T;
  while (T--) Main();
}