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>
// #include <vector>

#define repo(i,n) for (int i = 0; i < n; i++)
using namespace std;
using ll = long long;
struct UnionFind {
  vector<int> d;
  UnionFind(int n=0): d(n,-1) {}
  int find(int x) {
    if (d[x] < 0) return x;
    return d[x] = find(d[x]);
  }
  bool unite(int x, int y) {
    x = find(x); y = find(y);
    if (x == y) return false;
    if (d[x] > d[y]) swap(x,y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(int x, int y) { return find(x) == find(y);}
  int size(int x) { return -d[find(x)];}
};


    int deg[100005];
    vector<int> to[100005];
int main(){
    
    vector<int> d(4,-1);
    int n,m,k;
    cin >> n>> m >> k;

    UnionFind uf(n);

    repo(i,m){
        int a,b;
        cin >> a >> b;
        --a;
        --b;
        deg[a]++;
        deg[b]++;
        uf.unite(a,b);
        
//         repo(j,n){
//             cout << deg[j] << " ";
//         }
//         cout << endl;
//
//
//         repo(j,n){
//             cout << uf.size(j) << " ";
//         }
//         cout << endl;
    }
    repo(i,k){
        int a,b;
        cin >> a>> b;
        --a;
        --b;
        to[a].push_back(b);
        to[b].push_back(a);

    }
    repo(i,n){
        int ans = uf.size(i) -1 -deg[i];
//         cout << i << "の答えは" << ans << endl;
        for(int u : to[i]){
            if(uf.same(i,u)){
                --ans;
//                 cout << i << "と" << u << "はおなじunateでした よってansは" << ans << endl;
            }
            
        }
                
//         cout << ans;
//         if(i < n-1) cout << " ";
//         else cout << endl;
        printf("%d%c", ans, i==n-1?'\n':' ');
    }
//     cout << endl;

    return 0;
}