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;

// 人を頂点とし，友達関係を辺とするグラフにおいて，連結成分のサイズから友達||ブロック関係を除く

struct UnionFind
{
    vector<int> d;
    UnionFind(int n) : d(n, -1) {}
    int root(int x)
    {
        if (d[x] < 0)
            return x;
        return d[x] = root(d[x]);
    }
    bool unite(int x, int y)
    {
        x = root(x);
        y = root(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 root(x) == root(y); }
    int size(int x) { return -d[root(x)]; }
};

int main()
{
    int N, M, K;
    cin >> N >> M >> K;
    UnionFind graph(N);
    vector<int> friendship(N, 0);
    vector<vector<int>> block(N);
    for (int i = 0; i < M; i++)
    {
        int a, b;
        cin >> a >> b;
        a--, b--;
        friendship[a]++;
        friendship[b]++;
        graph.unite(a, b);
    }
    for (int i = 0; i < K; i++)
    {
        int a, b;
        cin >> a >> b;
        a--, b--;
        block[a].push_back(b);
        block[b].push_back(a);
    }

    for (int i = 0; i < N; i++)
    {
        int ans = graph.size(i) - friendship[i] - 1;
        for (int u : block[i])
        {
            if (graph.same(i, u))
                ans--;
        }
        printf("%d%c", ans, i == N - 1 ? '\n' : ' ');
    }
    return 0;
}