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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

vector<vector<int>> friends;
vector<vector<int>> blocks;

void solve(const int cur, const int current_turn, vector<int> &acc, vector<int> &visited, const int N) {
    if (visited[cur] > 0) return;
    visited[cur] = current_turn;
    acc.push_back(cur);
    for (auto &next : friends[cur]) {
        if (visited[next] == 0) {
            solve(next, current_turn, acc, visited, N);
        }
    }
}

int main(void) {
    int N, M, K;
    cin >> N >> M >> K;

    friends.resize(N);
    for (int i = 0; i < M; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        friends[a].push_back(b);
        friends[b].push_back(a);
    }

    blocks.resize(N);
    for (int i = 0; i < K; i++) {
        int c, d;
        cin >> c >> d;
        c--;
        d--;
        blocks[c].push_back(d);
        blocks[d].push_back(c);
    }

    vector<int> ans(N, 0);
    vector<int> visited(N, 0);
    int current_turn = 0;
    for (int i = 0; i < N; i++) {
        if (visited[i] == 0) {
            current_turn++;
            vector<int> visited_vs;
            solve(i, current_turn, visited_vs, visited, N);
            for (auto &v : visited_vs) {
                ans[v] = (int) visited_vs.size();
            }
        }
    }

    for (int i = 0; i < N; i++) {
        ans[i]--; // remove myself
        ans[i] -= friends[i].size(); // already friends
        for (auto &v : blocks[i]) {
            if (visited[v] == visited[i]) {
                ans[i]--;
            }
        }
    }

    cout << ans[0];
    for (int i = 1; i < N; i++) {
        cout << " " << ans[i];
    }
    cout << endl;

    return 0;
}