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:
    parent = []
    separate = []
    count = []

    def Create(self, N):
        for i in range(N+1):
            self.parent.append(i)
            self.count.append(1)
            self.separate.append(0)

    def Root(self, i):
        if self.parent[i] == i:
            return i
        else:
            return self.Root(self.parent[i])

    def Unite(self, x, y):
        rx = self.Root(x)
        ry = self.Root(y)
        if rx != ry:
            self.parent[rx] = ry
            self.count[ry] += self.count[rx]

    def Same(self, x, y):
        if self.Root(x) == self.Root(y):
            return True
        else:
            return False

    def Devide(self, x, y):
        if self.Same(x, y):
            self.separate[x] += 1
            self.separate[y] += 1

    def Count(self, x):
        c = self.count[self.Root(x)] - 1
        c -= self.separate[x]
        return c


def main():
    n,m,k = list(map(int, input().split()))
    uf = UnionFind()
    uf.Create(n)
    friend = [[] for i in range(n+1)]
    block = [[] for i in range(n+1)]
    ans = [0 for i in range(n+1)]

    for i in range(m):
        a,b = map(int, input().split())
        friend[a].append(b)
        friend[b].append(a)
        uf.Unite(a,b)
    
    for i in range(k):
        c,d = map(int, input().split())
        uf.Devide(c,d)
    
    for i in range(1,n+1):
        ans[i] = uf.Count(i) - len(friend[i])

    for i in range(1,n+1):
        print(ans[i],end="")
        if i != n:
            print(" ",end="")

    print()
main()