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 <bits/stdc++.h>

using namespace std;
using i64 = int64_t;
// #define int i64
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
const int INF = (1 << 30);
const i64 INFL = (1LL << 62);
const i64 MOD = 1000000007;

struct UnionFindSize {
  std::vector<int> data;
  const int n;
  UnionFindSize(int num) : data(n, -1), n(num) { data.resize(n); }
  int find(int k) {
    assert(0 <= k && k < n);
    if (data[k] < 0) return k;
    return (data[k] = find(data[k]));
  }
  bool unite(int x, int y) {
    assert(0 <= x && x < n);
    assert(0 <= y && y < n);
    x = find(x), y = find(y);
    if (x == y) return false;
    if (data[x] > data[y]) std::swap(x, y);
    data[x] += data[y];
    data[y] = x;
    return true;
  }
  int size(int k) {
    assert(0 <= k && k < n);
    return (-data[find(k)]);
  }
  bool same(int x, int y) {
    assert(0 <= x && x < n);
    assert(0 <= y && y < n);
    return find(x) == find(y);
  }
};

void print() { std::cout << std::endl; }
template <typename T, typename... A>
void print(const T& first, const A&... rest) {
  cout << sizeof...(rest) << endl;
  std::cout << first;
  if (sizeof...(rest)) std::cout << " ";
  print(rest...);
}
template <typename... A>
void print(const A&... rest) {
  print(rest...);
}
template <typename A>
void print(const std::vector<A>& v) {
  std::for_each(v.begin(), v.end(), [](A x) { std::cout << x << " "; });
  std::cout << std::endl;
}
// #undef int
int main() {
  // #define int i64
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  UnionFindSize uf(n);
  vector<vector<int>> gf(n);
  vector<vector<int>> gb(n);
  rep(i, m) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    uf.unite(a, b);
    gf[a].push_back(b);
    gf[b].push_back(a);
  }
  vector<int> d(n, 0);
  rep(i, k) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    gb[a].push_back(b);
    gb[b].push_back(a);
    if (uf.same(a, b)) {
      d[a]++;
      d[b]++;
    }
  }
  vector<int> ans(n);
  rep(i, n) { ans[i] = max(0, uf.size(i) - 1 - (int)(gf[i].size()) - d[i]); }
  rep(i, n) {
    cout << ans[i];
    if (i != n) {
      cout << " ";
    }
  }
  cout << endl;
  // cout << "f" << endl;

  // rep(i, n) { print(gf[i]); }
  // cout << "block" << endl;
  // rep(i, n) { print(gb[i]); }
  // print(uf.data);
  return 0;
}