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>

#define fst(t) std::get<0>(t)
#define snd(t) std::get<1>(t)
#define thd(t) std::get<2>(t)
#define unless(p) if(!(p))
#define until(p) while(!(p))

using ll = std::int64_t;
using P = std::tuple<int,int>;

struct UnionFind{
    std::vector<int> parent, rank, cnt;

    UnionFind(int n) : parent(n), rank(n, 0), cnt(n, 1) {
        std::iota(parent.begin(), parent.end(), 0);
    }

    int find(int v){
        if(v == parent[v]){return v;}
        return parent[v] = find(parent[v]);
    }

    bool same(int v, int w){
        return find(v) == find(w);
    }

    void unite(int v, int w){
        v = find(v);
        w = find(w);

        if(v == w){return;}

        if(rank[v] > rank[w]){
            parent[w] = v;
            cnt[v] += cnt[w];
        }else{
            parent[v] = w;
            cnt[w] += cnt[v];
            if(rank[v] == rank[w]){rank[w] += 1;}
        }
    }

    int count(int v){
        return cnt[find(v)];
    }
};

int N, M, K;
std::vector<int> G[100100], H[100100];
UnionFind uf(100100);

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

    std::cin >> N >> M >> K;

    for(int i=0;i<M;++i){
        int a, b;
        std::cin >> a >> b;

        G[a].emplace_back(b);
        G[b].emplace_back(a);

        uf.unite(a, b);
    }

    for(int i=0;i<K;++i){
        int a, b;
        std::cin >> a >> b;

        H[a].emplace_back(b);
        H[b].emplace_back(a);        
    }

    for(int i=1;i<=N;++i){
        int n = uf.count(i), m = 0, l = 0;

        for(int j : G[i]){
            if(uf.same(j, i)){
                m += 1;
            }
        }

        for(int j : H[i]){
            if(uf.same(j, i)){
                l += 1;
            }
        }

        int res = (n - 1) - m - l;

        std::cout << res << " \n"[i == N];
    }
}