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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

struct UnionFind {
    vector<long long> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(long long N) : par(N) { //最初は全てが根であるとして初期化
        for(long long i = 0; i < N; i++) par[i] = i;
    }

    long long  root(long long  x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(long long x, long long y) { // xとyの木を併合
        long long rx = root(x); //xの根をrx
        long long ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(long long x, long long y) { // 2つのデータx, yが属する木が同じならtrueを返す
        long long rx = root(x);
        long long ry = root(y);
        return rx == ry;
    }
};

int main() {
    long long n, m, k;
    cin >> n >> m >> k;
    long long a[m], b[m]; 
    long long c[k], d[k]; 
    long long ans[n][n];
    long long sum;

    for (long long i = 0; i < n; i++){
        for (long long j = 0; j < n; j++){
            ans[i][j]= 1;
        }
    }
    UnionFind tree(n+1);
    //cout << n << " " << m << " " << k << endl;
    for (int i = 0; i < m; i++){
        cin >> a[i] >> b[i];
        tree.unite(a[i], b[i]);

        if (a[i] <= b[i]){ 
            ans[a[i]-1][b[i]-1] = 0;
        }
        else{
            ans[b[i]-1][a[i]-1] = 0;
        }
    }
    for (long long i = 0; i < k; i++){
        cin >> c[i] >> d[i];
        if (c[i] <= d[i]){ 
            ans[c[i]-1][d[i]-1] = 0;
        }
        else{
            ans[d[i]-1][c[i]-1] = 0;
        }
    }

    /*
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            cout << ans[i][j] << ' ';
        }
        cout << endl;
    }
    */


    for (long long i = 0; i < n; i++){
        sum = 0;
        for (long long j = i+1; j < n; j++){
            if (tree.same(i+1, j+1)){
                sum += ans[i][j];
            }
            //cout << ans[i][j] << ' ';
        }
        //cout <<"row sum:" << sum << endl;
        for (int j = 0; j < i; j++){
            if (tree.same(i+1, j+1)){
                sum += ans[j][i];
            }
            //cout << ans[i][j] << ' ';
        }
        //cout << endl;
        cout << sum << ' ';
    }
    /*
    cout << endl;
    for (int i = 1; i <= n; i++){
        cout << i << " | ";
        for (int j = 1; j <= n; j++){
            cout << tree.same(i, j) << " ";
        }
        cout << endl;
    }
    */
    return 0;
}