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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

void dfs(int, int, vector<vector<bool>>&);

int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<bool>> friends(n, vector<bool>(n, false)), blocks(n, vector<bool>(n, false)), prospects(n, vector<bool>(n, false));
  //cout << "A" << endl;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    friends[a-1][b-1] = true;
    friends[b-1][a-1] = true;
    prospects[a-1][b-1] = true;
    prospects[b-1][a-1] = true;
  }
  //cout << "B" << endl;
  for (int i = 0; i < k; i++) {
    int c, d;
    cin >> c >> d;
    blocks[c-1][d-1] = true;
    blocks[d-1][c-1] = true;
  }
  //cout << "C" << endl;
  for (int i = 0; i < n; i++) {
    prospects[i][i] = true;
  }

  //cout << "input: comp" << endl;
  
  for (int p = 0; p < n; p++) {
    dfs(p, p, prospects);
  }

  for (int p = 0; p < n; p++) {
    int friend_pros = -1; //自分
    for (int q = 0; q < n; q++) {
      friend_pros += (int)((prospects[p][q]) && (!friends[p][q]) && (!blocks[p][q]));
    }

    cout << friend_pros << " ";
  }
  cout << endl;
}

void dfs(int x, int y, vector<vector<bool>> &prs) { //xの人から初めて行けるだけ行く（今yの人につながっている）
  //cout << x << ", " << y << endl;
  prs[x][y] = true;
  prs[y][x] = true;
  //たどり着けたので友達

  for (int i = 0; i < prs.size(); i++) {
    if ((x != y) && prs[y][i] && !prs[x][i]) { //x!=yのとき）xとiはつながっていないがxとy, yとiはつながっている
      dfs(x, i, prs);
    }
    else if ((x == y) && prs[x][i] && (x != i)) { //スタートの時はつながっているひと全員に
      dfs(x, i, prs);
    }
  }
  return ;
}