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;
using ll = long long;
using ld = long double;

vector<int> seen;
void dfs(vector<vector<int>> g, int v, int u){
    seen[v]=u;
    for(int i = 0; i < g[v].size(); i++) {
        if(seen[g[v][i]]==-1){
            dfs(g,g[v][i],u);
        }
    }
}

//#define int long long
//signed main(){
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n,m,k;
    cin >> n >> m >> k;
    int a,b;
    vector<vector<int>> g(n);
    for(int i = 0; i < m; i++) {
        cin >> a >> b;
        --a;
        --b;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }

    seen.assign(n,-1);
    int u=0;
    int idx=0;
    for(int i = 0; i < n; i++) {
        if(seen[i]==-1){
            dfs(g,i,u);
            u++;
        }
    }

    int un[u]={};
    for(int i = 0; i < n; i++) {
        un[seen[i]]++;
    }

    int ans[n];
    for(int i = 0; i < n; i++) {
        ans[i]=un[seen[i]]-g[i].size()-1;
    }

    int c,d;
    for(int i = 0; i < k; i++) {
        cin >> c >> d;
        --c;
        --d;
        if(seen[c]==seen[d]){
            ans[c]--;
            ans[d]--;
        }
    }

    for(int i = 0; i < n; i++) {
        cout << ans[i] << " ";
    }
    cout << "\n";

    return 0;
}