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>
using namespace std;
vector<int>vect[100001];
vector<int>block[100001];
set<int>r;

void dfs(bool visited[],int src)
{
       r.insert(src);
      visited[src]=true;

       for(int i=0;i<vect[src].size();i++)
       {
           if(visited[vect[src][i]]==false)
                dfs(visited,vect[src][i]);
       }
 }
int main()
{
    int n,m,k;
     cin>>n>>m>>k;

       for(int i=1;i<=m;i++)
       {
        int u,v;
         cin>>u>>v;

           vect[u].push_back(v);
           vect[v].push_back(u);
       }
       for(int i=1;i<=k;i++)
        {
           int u,v;
            cin>>u>>v;

            block[u].push_back(v);
            block[v].push_back(u);
        }

        bool visited[n+1]={false};

        int arr[n+1];

    for(int i=1;i<=n;i++)
    {
       if(visited[i]==false)
       {
          r.clear();
         dfs(visited,i);

          for(auto itr = r.begin();itr!=r.end();itr++)
          {
            int x = (*itr);

             arr[x] = r.size()-vect[x].size()-1;

               for(int j=0;j<block[x].size();j++)
                 if(r.find(block[x][j])!=r.end()) arr[x]--;
           }
       }
    }
      for(int i=1;i<=n;i++)
        cout<<arr[i]<<" ";
}