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: Time Limit Exceeded)

class UnionFind:  # O(α(N))
    def __init__(self, size):  # construct a Union-Find tree (1-idx)
        self.parent = [i for i in range(size)]
        self.rank = [0] * size
 
    def find(self, x):  # find the group (root) of a vertex
        if self.parent[x] == x:
            return x
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
 
    def is_same(self, x, y):  # check two vertices are in the same group
        return self.find(x) == self.find(y)
 
    def unite(self, x, y):  # unite two groups
        x, y = self.find(x), self.find(y)
        if x == y:  # the same group
            return
        # unite a small one to a bigger one to balance trees
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1


def main():
    N,M,K = map(int,input().split())
    fris = set()
    for _ in range(M):
        fris.add(tuple(sorted([int(i) -1 for i in input().split()])))
    blos = set()
    for _ in range(K):
        blos.add(tuple(sorted([int(i) -1 for i in input().split()])))
    uf = UnionFind(N)
    for a,b in fris:
        #if a == b-1 or b == a-1:
        uf.unite(a,b)
    alls = fris.union(blos)
    #print(alls)
    countl = []
    for i in range(N):
        count = 0
        for j in range(N):
            if i == j:
                #print(i,"==",j)
                continue
            if tuple(sorted([i, j])) in alls:
                #print(i,"fri or blo",j)
                continue
            #if i == j -1 or j == i-1:
            #    count += 1
            if uf.is_same(i,j):
                count += 1
            #else:
            #    print(i,"else",j)
        countl.append(str(count))
    print(" ".join(countl))

    #print("Yes") if uf.is_same(a, b) else print("No")
    
                

if __name__=="__main__":
    main()