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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;

VVI friends;
VVI blocks;

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

int main() {
  int N, M, K;
  cin >> N >> M >> K;
  UnionFind uf(N);
  friends.resize(N);
  rep(mi, M) {
    int A, B;
    cin >> A >> B;
    A--;
    B--;
    friends.at(A).push_back(B);
    friends.at(B).push_back(A);
    uf.unite(A, B);
  }
  blocks.resize(N);
  rep(ki, K) {
    int C, D;
    cin >> C >> D;
    C--;
    D--;
    blocks.at(C).push_back(D);
    blocks.at(D).push_back(C);
  }
  rep(ni, N) {
    int ans = uf.size(ni) - friends.at(ni).size();
    ans--;
    for (int b : blocks.at(ni)) {
      if (uf.same(ni, b)) {
        ans--;
      }
    }
    if (ni != N - 1) {
      cout << ans << " ";
    } else {
      cout << ans << endl;
    }
  }
  return 0;
}