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 ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep(i, n) for(ll i = 0;i < n;i++)
#define all(i) i.begin(), i.end()
template<class T, class U> bool cmax(T& a, U b) { if (a<b) {a = b; return true;} else return false; }
template<class T, class U> bool cmin(T& a, U b) { if (a>b) {a = b; return true;} else return false; }

class UnionFind {
private:
  vector<ll> parent;
  vector<ll> rank;

public:
  vector<ll> sz;
  UnionFind(ll size) : parent(size), rank(size, 0), sz(size, 1) {
    for (ll i = 0; i < size; i++) {
      parent[i] = i;
    }
  }

  ll find(ll x) {
    if (parent[x] == x) {
      return x;
    } else {
      return parent[x] = find(parent[x]);
    }
  }

  void unite(ll x, ll y) {
    x = find(x);
    y = find(y);
    if (x == y) return;

    if (rank[x] < rank[y]) {
      parent[x] = y;
      sz[y] += sz[x];
    } else {
      parent[y] = x;
      sz[x] += sz[y];
      if (rank[x] == rank[y]) rank[x]++;
    }
  }

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

int main() {
  cin.tie(0); ios::sync_with_stdio(false);

  ll n, m, k;
  cin >> n >> m >> k;
  vl a(m), b(m), c(k), d(k);
  vector<vl> block(n), tomo(n);

  UnionFind uf(n);

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

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

  using pll = pair<ll, ll>;
  vector<vector<pll>> edge(n);
  rep(i, m) {
    if (uf.same(a[i], b[i])) {
      edge[uf.find(a[i])].emplace_back(a[i], b[i]);
    }
  }
  rep(i, k) {
    if (uf.same(c[i], d[i])) {
      edge[uf.find(c[i])].emplace_back(c[i], d[i]);
    }
  }

  vl ans(n);
  rep(i, n) {
    ans[i] = uf.sz[uf.find(i)]-1;
  }

  rep(i, n) {
    for (auto&& j : edge[i]) {
      ans[j.first]--;
      ans[j.second]--;
    }
  }

  rep(i, n) {
    cout << ans[i];
    if (i == n-1) cout << endl;
    else cout << " ";
  }
}