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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i, n) for(long long i = 0; i < (long long)(n); i++)
using namespace std;
vector<vector<int> > f;
vector<vector<int> > b;
vector<int> g;
int n,m,k;

void dfs(int i, int num){
    if(i<n && g.at(i)==-1){
        g.at(i)=num;
        rep(j,f.at(i).size())dfs(f.at(i).at(j),num);
    }
}

int main(){
    cin >> n >> m >> k;
    rep(i,n){
        f.emplace_back(vector<int>());
        b.emplace_back(vector<int>());
        g.emplace_back(-1);
    }
    rep(i,m){
        int a,bb;
        cin >> a >> bb;
        a--;
        bb--;
        f.at(a).emplace_back(bb);
        f.at(bb).emplace_back(a);
    }
    rep(i,k){
        int c,d;
        cin >> c >> d;
        c--;
        d--;
        b.at(c).emplace_back(d);
        b.at(d).emplace_back(c);
    }
    rep(i,n)dfs(i,i);
    rep(i,n){
        int ans;
        ans = 0;
        rep(j,n){
            if(g.at(i) == g.at(j)){
                ans++;
                bool flag;
                flag = false;
                rep(k,f.at(i).size())if(f.at(i).at(k) == j){
                    flag = true;
                    break;
                }
                if(!flag){
                    rep(k,b.at(i).size())if(b.at(i).at(k) == j){
                        flag = true;
                        break;
                    }
                }
                if(flag)ans--;
            }
        }
        ans--;
        cout << ans;
        if(i != n-1)cout << " ";
    }
    cout << endl;
}