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;
  #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  #define endl "\n"
  #define all(v) v.begin(),v.end()
  #define f first
  #define s second
  #define PB push_back
  #define MP make_pair
  #define double long double
  #define trace1(x)                cerr<<#x<<": "<<x<<endl
  #define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
  #define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
  #define trace4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
  #define trace5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
  #define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
  #define cases int testcases;cin>>testcases; while(testcases--)
  #define MOD 998244353
  #define N 100005
  #define INF 1000000000000
  #define eps 1e-6
  #define int long long



int parent[N],sz[N];
void make_set(int n)
{
  for(int i = 0; i <= n; i++)
  {
    parent[i] = i;
    sz[i] = 1;
  }
}

int getParent(int s)
{
  if(s == parent[s])return s;
  else return parent[s] = getParent(parent[s]);
}

void join(int a,int b)
{
  a = getParent(a);
  b = getParent(b);

  if(a != b)
  {
    parent[b] = a;
    sz[a] += sz[b];
    sz[b] = 0;
  }
}

int ans1[N];
int ans2[N];

int32_t main()
{



    int n,m,k;
    cin >> n >> m >> k;

    make_set(n);

     for(int i = 1; i <= m; i++)
     {
      int x,y;
      cin >> x >> y;
     // trace2(sz[x],sz[y]);
      join(x,y);
      ans1[x]++;
      ans1[y]++;

     }

     for(int i = 1; i <= k; i++)
     {
      int x,y;
      cin >> x >> y;
      if(getParent(x) == getParent(y))
      {
        ans2[x]++;
        ans2[y]++;
      }
     }

     for(int i = 1; i <= n; i++)
     {
      int y = getParent(i);
      cout << sz[y] - ans1[i] - ans2[i] - 1 << " ";
     }



    return 0;
}