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: Wrong Answer)

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <regex>
#include <numeric>
using namespace std;

/*--- Macro ---*/
#define rep(i, n) for(int i=0; i<(int)n; i++)
#define rep1(i, n) for(int i=1; i<(int)n; i++)
#define mp make_pair
#define pb push_back
#define all(a) (a).begin(), (a).end()

/*---  Type ---*/
typedef long long ll;
typedef pair<int, int> P;

/*--- Constants ---*/
template <typename T> const T INF = numeric_limits<T>::max() / 10;
const int MOD = (int)1e9 + 7;

// Floating
const double EPS = 1e-11;
#define EQ0(x) (abs(x) < EPS)
#define EQ(x, y) (abs(a-b) < EPS)

// Util
int max(int a, int b, int c) {return max(a, max(b, c));}
int min(int a, int b, int c) {return min(a, min(b, c));}
ll max(ll x, ll y) {return x > y ? x : y;}
ll min(ll x, ll y) {return x < y ? x : y;}
ll max(ll a, ll b, ll c) {return max(a, max(b, c));}
ll min(ll a, ll b, ll c) {return min(a, min(b, c));}

// How to use 「map」
/*
  - 宣言
    map<{key_type}, {value_type}> salary;
  - 要素の挿入
    salary["Alice"] = 1000;
  - 要素の取得 O(logn)
    salary.at("Alice");
  - 要素の削除
    salary.erase("Alice");
  - キー検索
    salary.find("Alice"); -> true
    salary.find("Bob");   -> false
  - イテレーション
    for(auto& [key, value] : salary) {...}
 */
// -------------------------------------


// How to use 「lower_bound」, 「upper_bound」
/*
  vector<int> a = {1, 2, 3, 4, 5};
  auto iter1 = lower_bound(all(a), 3);  <- 第二引数「以上」の最小イテレータが返る
  auto iter2 = upper_bound(all(a), 3);  <- 第二引数「より大きい」最小のイテレータが返る
 */
// -------------------------------------


// How to use 「struct sort」
/*
  struct Person {
    string name;
    int age;

    Person(string _name, int _age) {
      name = _name;
      age = _age;
    }

    bool operator<(const Person &another) const {
      return age < another.age;
    }
  };

  vector<Person> people;
  people.pb(Person("Alice", 20));
  people.pb(Person("Bob", 19));
  people.pb(Person("John", 40));
  people.pb(Person("Chris", 30));
  sort(all(people));
 */
 // -------------------------------------

// -------------------------------------------------------------------------------------

struct UnionFind {
  vector<int> parent;

  UnionFind(int N) : parent(N) {
    for(int i = 0; i < N; i++) parent[i] = i;
  }

  /**
    要素xの根を求める
    */
  int root(int x) {
    if(parent[x] == x) return x;
    return parent[x] = root(parent[x]);  // 経路圧縮しつつ根を返す
  }

  /**
    要素xと要素yの木を併合
    */
  void unite(int x, int y) {
    int x_root = root(x);
    int y_root = root(y);
    if(x_root == y_root) return;
    parent[x_root] = y_root;
  }

  /**
    要素xと要素yが同じ木に属しているか
    */
  bool same(int x, int y) {
    int x_root = root(x);
    int y_root = root(y);
    return x_root == y_root;
  }
};

int main() {
  int n, m, k;
  cin >> n >> m >> k;

  vector<int> follows(n, 0);

  UnionFind u = UnionFind(n);
  rep(i, m) {
    int a, b;
    cin >> a >> b;
    u.unite(a-1, b-1);
    follows[a-1]++;
    follows[b-1]++;
  }

  UnionFind u2 = UnionFind(n);
  rep(i, k) {
    int c, d;
    cin >> c >> d;
    u2.unite(c-1, d-1);
  }

  rep(i, n) {
    int cnt = 0;
    rep(j, n) {
      if (u.same(i, j) && !u2.same(i, j)) {
        cnt++;
      }
    }
    cout << cnt - follows[i];
    if (i != n-1) {
      cout << " ";
    } else {
      cout << endl;
    }
  }

  return 0;
}