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>
using namespace std;
#define ll  long long
#define endl '\n'
#define FASTINOUT ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
const ll MOD = 1000000007;
const double PI = 3.141592653589793238463;
const ll N=1e5+1;
set<int>fr[N];
set<int>en[N];
ll ans[N];
bool visited[N];
vector<int>kl;
void dfs(int x)
{
    visited[x]=true;
    for (auto i:kl){
        if (fr[i].find(x)==fr[i].end()&&en[i].find(x)==en[i].end()){
            ans[i]++;
            ans[x]++;
        }
    }
    kl.push_back(x);
    for (auto i=fr[x].begin();i!=fr[x].end();i++){
        if (!visited[*i])
            dfs(*i);
    }
}
int main()
{
    FASTINOUT;
    int x,m,k;
    cin>>x>>m>>k;
    int l,r;
    for (int i=0; i<m; i++)
    {
        cin>>l>>r;
        fr[l].insert(r);
        fr[r].insert(l);
    }
    for (int i=0; i<k; i++)
    {
        cin>>l>>r;
        en[l].insert(r);
        en[r].insert(l);
    }
    for (int i=1;i<=x;i++){
        if (!visited[i]){
            kl.clear();
            dfs(i);
        }
    }
    for (int i=1;i<=x;i++)
        cout<<ans[i]<<" ";
    return 0;
}