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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

// inputに入力データ全体が入る
function Main(input) {
  input = input.split("\n");
  tmp = input.shift();
  N = parseInt(tmp.split(" ")[0]);
  M = parseInt(tmp.split(" ")[1]);
  K = parseInt(tmp.split(" ")[2]);
  Fbj = {};
  Bbj = {};
  F = [];
  B = [];
  Hbj = {};
  kouho = {};
  kouho2 = {};
  kouho3 = [];
  for (i = 0; i < M; i++) {
    F.push(input[i]);
  }
  for (i = M; i < M + K; i++) {
    B.push(input[i]);
  }
  F.forEach(v => {
    v1 = parseInt(v.split(" ")[0]);
    v2 = parseInt(v.split(" ")[1]);
    Fbj["H" + v1] || (Fbj["H" + v1] = []);
    Fbj["H" + v1].push(v2);
    Fbj["H" + v2] || (Fbj["H" + v2] = []);
    Fbj["H" + v2].push(v1);
    Fbj["H" + v1] = Fbj["H" + v1].filter((x, i, self) => self.indexOf(x) === i);
    Fbj["H" + v2] = Fbj["H" + v2].filter((x, i, self) => self.indexOf(x) === i);
  });
  B.forEach(v => {
    v1 = parseInt(v.split(" ")[0]);
    v2 = parseInt(v.split(" ")[1]);
    Bbj["H" + v1] || (Bbj["H" + v1] = []);
    Bbj["H" + v1].push(v2);
    Bbj["H" + v2] || (Bbj["H" + v2] = []);
    Bbj["H" + v2].push(v1);
    Bbj["H" + v1] = Bbj["H" + v1].filter((x, i, self) => self.indexOf(x) === i);
    Bbj["H" + v2] = Bbj["H" + v2].filter((x, i, self) => self.indexOf(x) === i);
  });

  for (i = 1; i <= N; i++) {
    Hbj["H" + i] = [...Array(N).keys()].map(i => ++i).filter(v => v != i);
    Hbj["H" + i].forEach(v => {
      kouho["H" + i] || (kouho["H" + i] = []);
      if (Fbj["H" + i] && Fbj["H" + i].find(a => v == a)) return;
      if (Bbj["H" + i] && Bbj["H" + i].find(a => v == a)) return;
      kouho["H" + i].push(v);
    });
  }

  for (i = 1; i <= N; i++) {
    kouho2["H" + i] || (kouho2["H" + i] = []);
    kouho["H" + i].forEach(v => {
      if (kouho["H" + v].find(a => i == a)) {
        kouho2["H" + i].push(v);
      }
    });
  }

  for (i = 1; i <= N; i++) {
    kouho3[i - 1] || (kouho3[i - 1] = []);
    kouho2["H" + i].forEach(v => {
      if (Fbj["H" + v]) {
        // 友達候補
        Fbj["H" + v].forEach(w => {
          // 友達候補の友達
          if (Fbj["H" + w].find(x => i == x)) {
            kouho3[i - 1].push(v);
          }
        });
      }
    });
  }

  console.log(kouho3.map(v => v.length).join(" "));
}

//*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));