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 C (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
typedef struct{int Parent,Rank;}UnionFind;
static inline int Root(int x,UnionFind *uf)
{
  if(uf[x].Parent==x)return x;
  else return uf[x].Parent=Root(uf[x].Parent,uf);
}
static inline int Unite(int x,int y,UnionFind *uf)
{
  x=Root(x,uf),y=Root(y,uf);
  if(x==y)return 0;
  uf[x].Parent=y;uf[y].Rank+=uf[x].Rank;return 1;
}
static inline int Size(int x,UnionFind *uf){return uf[Root(x,uf)].Rank;}
static inline int Belong(int x,int y,UnionFind *uf){return Root(x,uf)==Root(y,uf);}
int main(void)
{
  int A,B,C,D,N,M,K;
  scanf("%d %d %d",&N,&M,&K);
  int known[N];
  UnionFind uft[N];
  for(int i=0;i<N;i++){known[i]=0;uft[i].Parent=i;uft[i].Rank=1;}
  while(M--)
  {
    scanf("%d %d",&A,&B);
    A--,B--;known[A]++,known[B]++;Unite(A,B,uft);
  }
  while(K--)
  {
    scanf("%d %d",&C,&D);
    C--,D--;if(Belong(C,D,uft))known[C]++,known[D]++;
  }
  for(int i=0;i<N;i++){printf("%d ",Size(i,uft)-known[i]-1);}
  return printf("\n"),0;
}