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)

/**
 *    author:  zjsdut
 *    created: 2020/03/29 20:53:44
 **/
#include <bits/stdc++.h>

using namespace std;

struct fast_ios {
  fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
  };
} fast_ios_;

template <typename A, typename B> bool chkmin(A &a, const B &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}

template <typename A, typename B> bool chkmax(A &a, const B &b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}

template <typename T> istream &operator>>(istream &stream, vector<T> &vec) {
  for (auto &x : vec)
    stream >> x;
  return stream;
}

void scan() {}

template <class T, class... Args> void scan(T &a, Args &... rest) {
  cin >> a;
  scan(rest...);
}

using ll = long long;
using vl = vector<ll>;
using vb = vector<bool>;
using vi = vector<int>;
using pii = pair<int, int>;
template <typename T> using vv = vector<vector<T>>;
#define rng(i, a, b) for (int i = (int)(a); i < (int)(b); ++i)
#define up(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define down(i, b, a) for (int i = int(b); i >= int(a); i--)
#define TOKENPASTE(x, y) x##y
#define TOKENPASTE2(x, y) TOKENPASTE(x, y)
#define rep(n)                                                                 \
  for (int TOKENPASTE2(_iter_, __LINE__) = 0,                                  \
                               TOKENPASTE2(_num_, __LINE__) = (int)(n);        \
       TOKENPASTE2(_iter_, __LINE__) < TOKENPASTE2(_num_, __LINE__);           \
       ++TOKENPASTE2(_iter_, __LINE__))
#define FOR(x, cont) for (const auto &x : cont)
#define For(x, cont) for (auto &x : cont)
#define all(x) begin(x), end(x)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ep emplace
#define SZ(x) (int)(x).size()
#define UNIQ(vec) (vec).erase(unique(all(vec)), end(vec))
#define LB(cont, x) int(lower_bound(all(cont), x) - begin(cont))
#define UB(cont, x) int(upper_bound(all(cont), x) - begin(cont))

template <typename T>
ostream &operator<<(ostream &stream, const vector<T> &vec) {
  if (!vec.empty()) {
    stream << vec[0];
    for (size_t i = 1; i != vec.size(); ++i)
      stream << ' ' << vec[i];
  }
  return stream;
}

template <typename T> void print(const vector<T> &t) { cout << t << '\n'; }

template <typename T> void print(const vector<vector<T>> &t) {
  for (const auto &row : t) {
    print(row);
  }
}

template <typename T> void print(const T &t) { cout << t << ' '; }

template <typename T, typename... Args>
void print(const T &t, const Args &... rest) {
  print(t);
  print(rest...);
}

template <typename T> void println(const T &t) { cout << t << '\n'; }

template <typename T, typename... Args>
void println(const T &t, const Args &... rest) {
  print(t);
  println(rest...);
}
class UnionFind {
  vi par, sz;
  int nTree;

public:
  explicit UnionFind(int n) : par(n), sz(n, 1), nTree(n) { // 0-indexed
    iota(par.begin(), par.end(), 0);
  }

  int n_tree() const { return nTree; }

  int size(int x) { return sz[root(x)]; }

  int root(int x) { return x == par[x] ? x : par[x] = root(par[x]); }
  bool same(int x, int y) {
    return root(x) == root(y);
  }
  bool unite(int x, int y) {
    int rx = root(x), ry = root(y);
    if (rx != ry) {
      par[rx] = ry;
      --nTree;
      sz[ry] += sz[rx];
      return true;
    }
    return false;
  }

  bool is_root(int x) const { return par[x] == x; }
};
int main() {
#if defined LOCAL && !defined DUIPAI
  ifstream in("in.txt");
  cin.rdbuf(in.rdbuf());
#endif
  int n, m, k;
  scan(n, m, k);
  vv<int> B(n);
  vi cnt(n);
  UnionFind uf(n);
  rep (m) {
      int a,b;
      scan(a, b);
      --a, --b;
      uf.unite(a, b);
      cnt[a]++, cnt[b]++;
  }
  rep (k) {
    int a,b;
    scan(a, b);
    --a, --b;
    B[a].pb(b);
    B[b].pb(a);
  }
  rng (i, 0, n) {
    int t = uf.size(i) - cnt[i] - 1;
    FOR (j, B[i]) {
      t -= uf.same(i, j);
    }
    print(t);
  }
  return 0;
}