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 Common Lisp (SBCL 1.1.14) (judge verdict: Time Limit Exceeded)

(defvar *already-looked*)

(defun solve (n m k friends blocks)
  (declare (ignore m k))
  (let ((friends-maps
          (make-array n
                      :element-type 'hash-table
                      :initial-contents (loop repeat n collect (make-hash-table))))
        (blocks-maps
          (make-array n
                      :element-type 'hash-table
                      :initial-contents (loop repeat n collect (make-hash-table)))))
    (loop for (a . b) in friends
          do (setf (gethash b (aref friends-maps (1- a))) t
                   (gethash a (aref friends-maps (1- b))) t))
    (loop for (a . b) in blocks
          do (setf (gethash b (aref blocks-maps (1- a))) t
                   (gethash a (aref blocks-maps (1- b))) t))
    (flet ((friends-p (a b)
             (gethash b (aref friends-maps (1- a))))
           (blocks-p (a b)
             (gethash b (aref blocks-maps (1- a))))
           (friends-of (a)
             (loop for b being the hash-keys of (aref friends-maps (1- a))
                   collect b))
           (blocks-of (a)
             (loop for b being the hash-keys of (aref blocks-maps (1- a))
                   collect b)))
      (let ((counts (make-array n :element-type 'fixnum)))
        (labels ((main (a from)
                   (setf (gethash a *already-looked*) t)
                   (loop for b in (friends-of a)
                         unless (gethash b *already-looked*)
                         do
                         (setf (gethash b *already-looked*) t)
                         (when (and (not (= a from))
                                    (not (friends-p from b))
                                    (not (blocks-p from b)))
                           (incf (aref counts (1- from))))
                         (loop for c in (friends-of b)
                               if (and (not (gethash c *already-looked*))
                                       (not (friends-p from c)))
                               do (unless (blocks-p from c)
                                    (incf (aref counts (1- from))))
                               (main c from)))))
          (loop for a from 1 to n
                do (let ((*already-looked* (make-hash-table)))
                     (main a a))))
        (coerce counts 'list)))))

#-swank
(let* ((n (read))
       (m (read))
       (k (read))
       (friends (loop repeat m collect (cons (read) (read))))
       (blocks (loop repeat k collect (cons (read) (read)))))
  (format t "~{~A~^ ~}~%" (solve n m k friends blocks)))