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: Accepted)

#!python3

iim = lambda: map(int, input().rstrip().split())

from bisect import bisect

def resolve():
    def find(i):
        x = S[i]
        if x < 0:
            return i
        S[i] = find(x)
        return S[i]

    def join(i, j):
        ai = find(i)
        aj = find(j)
        diff = ai - aj
        if diff == 0:
            return
        elif diff > 0:
            ai, aj = aj, ai
        S[ai] += S[aj]
        S[aj] = ai
    def same(i, j):
        ai = find(i)
        aj = find(j)
        return ai == aj
    def size(i):
        return -S[find(i)]

    N, M, K = iim()
    S = [-1] * N
    T = [1] * N
    for i in range(M):
        ai, bi = iim()
        ai -= 1; bi -= 1
        join(ai, bi)
        T[ai] += 1
        T[bi] += 1
        #print(ai, bi, S)

    for i in range(K):
        ai, bi = iim()
        ai -= 1; bi -= 1
        if same(ai, bi):
            T[ai] += 1
            T[bi] += 1
    #print(S)
    #print(T)

    ans = [size(i) - T[i] for i in range(N)]
    print(*ans)


if __name__ == "__main__":
    resolve()