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 Python (3.4.3) (judge verdict: Wrong Answer)

slct='D'

# C
if slct=='C':
    N,M=map(int,input().split())
    cnd=[tuple(int(i) for i in input().split()) for j in range(M)]
    ans=-1
    for n in range(10**N):
        # print(n)
        flag=True
        n_str=str(n)
        if len(n_str)!= N:
            continue
        for s,c in cnd:
            if n_str[s-1]!=str(c):
                # print(n_str[s-1],c)
                flag=False
                break
        if flag:
            ans=n
            break
    print(ans)

# D
elif slct=='D':
    N,M,K=map(int,input().split())
    
    par=[-1 for _ in range(N)]
    # 負なら根でabsがその根の属する木のサイズ
    # 正ならその値が親

    def root(n):
        if par[n]==n:
            return n
        else:
            par[n]=root(par[n]) # 親の根に繋ぎなおす
            return par[n]
    
    def unite(m,n):
        x=root(m)
        y=root(n)
        if x==y: # 同じ木にいる
            pass
        if par[x]>par[y]: # yのサイズの方が大きい:swap
            x,y=y,x 
        par[x]+=par[y] # サイズの大きいxにくっつける: サイズを記録
        par[y]=x 

    def issame(m,n):
        return root(m)==root(n)
    
    def size(n): # nの属する木のsizeを取得
        return -par[root(n)]
    
    # iの求めたい値 = size(i) -直接の友達関係の数 -iの木上でブロック関係にある数 -1
    # 直接の友達関係ならお互いに同じ木にいる
    ans=[-1 for _ in range(N)] # 自分の分: -1
    
    for _ in range(M): #木達の構築
        A,B=map(int,input().split())
        unite(A-1,B-1)
        # 直接の友達関係を引く
        ans[A-1]-=1
        ans[B-1]-=1
    for _ in range(K): # ブロック関係にあるかどうか
        C,D=map(int,input().split())
        #同じ木でなければ答えに関係ない,同じ木にいるなら引く
        if issame(C-1,D-1):
            ans[C-1]-=1
            ans[D-1]-=1
    for i in range(N): 
        print(ans[i]+size(i))