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: Runtime Error)

#include "bits/stdc++.h"
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll =long long;
using P =pair<int,int>;

int main(){
    int n,m,k;
    cin >> n >> m >> k;
    vector <int> a(m),b(m),c(k),d(k);
    vector <vector <int>> tomo(n+1),bro(n+1); 
    rep(i,m) {
        cin >> a[i] >> b[i];
        tomo[a[i]].push_back(b[i]);
        tomo[b[i]].push_back(a[i]);
    }
    rep(i,k){
        cin >> c[i] >> d[i];
        bro[c[i]].push_back(d[i]);
        bro[d[i]].push_back(c[i]);
    }

    vector <int> wa;
    wa.push_back(0);
    int kau;
    vector <int> syo(n+1,0);
    queue <int> q;
    int syozo;
    syozo=0;
    int l;
    int qq;

    rep(i,n){
        ++syozo;
        if (syo[i+1]!=0){
            continue;
        }
        syo[i+1]=syozo;
        kau=1;
        q.push(i+1);
        while(q.size()!=0){
            qq=q.front();
            q.pop();
            l=tomo[qq].size();
            rep(j,l){
            if(syo[tomo[qq][j]]==0){
                syo[tomo[qq][j]]=syozo;
                q.push(tomo[qq][j]);
                ++kau;
            }
            }
        }
        wa.push_back(kau);
    }
    
    int ans;
    rep(i,n){
        ans=wa[syo[i+1]]-tomo[i+1].size()-1;
        rep(j,bro[i+1].size()){
            if(syo[i+1]==syo[bro[i+1][j]]){
                --ans;
            }
        }
        printf("%d ",ans);
    }
    

    return 0;
}