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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using P = pair<int, int>;
#define pb push_back


int N, M, K;
P f, b;
vector<P> friends(100010);
vector<int> f_degree(100010, 0);  //各頂点の次数
vector<P> block(100010);
vector<int> b_degree(100010, 0);  //同一グラフ内のブロック人数


struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) {}
    void init(int n) { par.assign(n, -1); }

    int root(int x){
        if(par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

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

    bool merge(int x, int y){
        x = root(x); y = root(y);
        if(x == y) return false;
        if(par[x] > par[y]) swap(x, y);
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x){
        return -par[root(x)];
    }

};

int main(){
    cin >> N >> M >> K;
    UnionFind uf(N+1);  //ユニオンファインド作成

    for(int i = 0; i < M; i++){ 
        cin >> f.first >> f.second; 
        friends[i] = f;
        ++f_degree[f.first]; ++f_degree[f.second];
        uf.merge(f.first, f.second);  //first-secondの辺を作成
    }

    for(int i = 0; i < K; i++){ 
        cin >> b.first >> b.second; 
        block[i] = b;
        if(uf.issame(b.first,b.second)){ ++b_degree[b.first]; ++b_degree[b.second]; }
    }

    /*iの友達候補=(i含グラフサイズ)-1-(i次数)-(i含グラフ内ブロック人数)*/
    for(int v = 1; v <= N; v++) cout << uf.size(v)-1-f_degree[v]-b_degree[v] << " ";
    cout << endl;

    return 0;

}