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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author aajisaka
 */

#include<bits/stdc++.h>

using namespace std;

void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define SPEED ios_base::sync_with_stdio(false);cin.tie(nullptr)
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define all(v) v.begin(), v.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

using ll = long long;
using P = pair<ll, ll>;

constexpr long double PI = 3.14159265358979323846264338327950288L;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

class DFriendSuggestions {
public:

    std::vector<int> par;
    std::vector<int> rak; // number of the vertices

    void init(int n) {
      par.resize(n);
      rak.resize(n);
      for(int i=0; i<n; i++) {
        par[i] = i;
        rak[i] = 1;
      }
    }

    int find(int x) {
      if (par[x] == x) {
        return x;
      } else {
        return par[x] = find(par[x]);
      }
    }

    void unite(int x, int y) {
      x = find(x);
      y = find(y);
      if (x==y) return;
      if (rak[x] < rak[y]) {
        rak[y] += rak[x];
        par[x] = y;
      } else {
        rak[x] += rak[y];
        par[y] = x;
      }
    }

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

    vector<int> blocked[100005];
    vector<int> fri[100005];
    vector<int> roots;

    void solve(istream& cin, ostream& cout) {
      SPEED;
      int n, m, k; cin >> n >> m >> k;
      init(n);

      rep(i, m) {
        int a, b; cin >> a >> b;
        a--; b--;
        unite(a, b);
        fri[a].push_back(b);
        fri[b].push_back(a);
      }

      rep(i, k) {
        int c, d; cin >> c >> d;
        c--; d--;
        blocked[c].push_back(d);
        blocked[d].push_back(c);
      }

      map<int, set<int>> mp;

      rep(i, n) {
        mp[find(i)].insert(i);
      }

      vector<int> ans(n);

      for(auto& e: mp) {
        set<int> st = e.second;
        for(auto f: e.second) {
          vector<int> erased;
          for(auto g: blocked[f]) {
            if (st.find(g) != st.end()) {
              st.erase(g);
              erased.push_back(g);
            }
          }
          for(auto g: fri[f]) {
            if (st.find(g) != st.end()) {
              st.erase(g);
              erased.push_back(g);
            }
          }
          ans[f] = st.size()-1;
          for(auto g: erased) {
            st.insert(g);
          }
        }
      }

      for(auto e: ans) {
        cout << e << ' ';
      }
      cout << endl;




    }
};

signed main() {
  DFriendSuggestions solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}