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)

/*  -*- coding: utf-8 -*-
 *
 * d.cc: D - Friend Suggestions
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;

/* constant */

const int MAX_N = 100000;

/* typedef */

typedef vector<int> vi;

struct UFT {
  int links[MAX_N], ranks[MAX_N], sizes[MAX_N];
  UFT() {}

  void init(int n) {
    for (int i = 0; i < n; i++) links[i] = i, ranks[i] = sizes[i] = 1;
  }

  int root(int i) {
    int i0 = i;
    while (links[i0] != i0) i0 = links[i0];
    return (links[i] = i0);
  }

  int rank(int i) { return ranks[root(i)]; }
  int size(int i) { return sizes[root(i)]; }
  bool same(int i, int j) { return root(i) == root(j); }

  int merge(int i0, int i1) {
    int r0 = root(i0), r1 = root(i1), mr;
    if (r0 == r1) return r0;
    if (ranks[r0] == ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      ranks[r0]++;
      mr = r0;
    }
    else if (ranks[r0] > ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      mr = r0;
    }
    else {
      links[r0] = r1;
      sizes[r1] += sizes[r0];
      mr = r1;
    }
    return mr;
  }
};

/* global variables */

UFT uft;
vi nbrs[MAX_N], blks[MAX_N];

/* subroutines */

/* main */

int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);

  uft.init(n);
  
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    nbrs[a].push_back(b);
    nbrs[b].push_back(a);
    uft.merge(a, b);
  }

  for (int i = 0; i < k; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    blks[a].push_back(b);
    blks[b].push_back(a);
  }

  for (int i = 0; i < n; i++) {
    int r = uft.root(i), sz = uft.size(i);
    vi &nbri = nbrs[i], &blki = blks[i];

    for (vi::iterator vit = blki.begin(); vit != blki.end(); vit++)
      if (uft.same(r, *vit)) sz--;
    sz -= nbri.size() + 1;

    if (i) putchar(' ');
    printf("%d", sz);
  }
  putchar('\n');
  return 0;
}