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<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#define mod 1000000007
typedef long long ll;
using namespace std;

class UnionFind {
public:
    vector <ll> par; 
    vector <ll> siz; 


    UnionFind(ll sz_): par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL); 
        for (ll i = 0; i < sz_; ++i) par[i] = i; 
    }

    
    ll root(ll x) {
        while (par[x] != x) {
            x = par[x] = par[par[x]]; 
        }
        return x;
    }

  
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;

        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y) { 
        return root(x) == root(y);
    }

    ll size(ll x) { 
        return siz[root(x)];
    }
};

int frenum[100005]; // 直接友達関係の数
int bro[100005]; // 同じ連結成分でブロック関係にあるが直接友達ではない人の数
// int fre[100005];
// int bro[100005];

int main(){
  int n,m,k; cin >> n >> m >> k;
  UnionFind tree(n);
  vector<set<int>> fre(n); // 人iと直接友達関係の人を記録
  for(int i=0;i<m;i++){
    int a,b; cin >> a >> b;
    a--; b--;
    frenum[a]++;
    frenum[b]++;
    tree.merge(a,b);
    fre[a].insert(b);
    fre[b].insert(a);
  }
  for(int i=0;i<k;i++){
    int c,d; cin >> c >> d;
    c--; d--;
    if(tree.issame(c,d)){
      // 連結成分内でブロック関係にあり、かつ直接友達でなければ重複することなく引くことができる
      // 直接友達でもあったならばここでは引く必要がない
      if(!(fre[c].count(d))) bro[c]++;
      if(!(fre[d].count(c))) bro[d]++;
    }
  }
  for(int i=0;i<n;i++){
    ll ans = tree.size(i) - frenum[i] - bro[i] - 1;
    if(i > 0) cout << " ";
    cout << ans;
  }
  cout << endl;
  return 0;
}