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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

class UnionFind():
    def __init__(self,n):
        self.n=n
        self.parents=[-1]*n
    def find(self,x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x]=self.find(self.parents[x])
            return self.parents[x]
    def unite(self,x,y):
        x = self.find(x)
        y = self.find(y)
        if x==y:
            return
        if self.parents[x]>self.parents[y]:
            x,y=y,x
        self.parents[x]+=self.parents[y]
        self.parents[y]=x
    def size(self, x):
        return -self.parents[self.find(x)]
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i)==root]
    def roots(self):
        return [i for i,x in enumerate(self.parents) if x<0]
    def group_count(self):
        return len(self.roots())
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}
n,m,k=map(int,input().split())
uf=UnionFind(n+1)
s=[set() for i in range(n+1)]
parent_list=set()
for i in range(m):
  a,b=map(int,input().split())
  s[a].add(b)
  s[b].add(a)
  uf.unite(a,b)
for i in range(k):
  c,d=map(int,input().split())
  s[c].add(d)
  s[d].add(c)
for i in range(1,n+1):
  parent_list.add(uf.find(i))
ans=[0]*(n+1)
for i in parent_list:
  l=set(uf.members(i))
  v=uf.size(i)
  for j in l:
    ans[j]-=len(l&s[j])
  for j in l:
    ans[j]+=v-1
print(*ans[1:])