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)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}

int tree[100000][3];
int root(int x){
	return tree[x][0]==x?x:(tree[x][0]=root(tree[x][0]));
}
void unon(int x, int y){
	x=root(x);
	y=root(y);
	if(x==y){
	}else if(tree[x][1]>tree[y][1]){
		tree[y][0]=x;
		tree[x][2]+=tree[y][2];
	}else if(tree[x][1]<tree[y][1]){
		tree[x][0]=y;
		tree[y][2]+=tree[x][2];
	}else{
		tree[y][0]=x;
		tree[x][1]++;
		tree[x][2]+=tree[y][2];
	}
}
int ans[100000];

int main(void){
	int n, m, k, i, j;
	int a[100000][2];
	scanf("%d%d%d", &n, &m, &k);
	for(i=0; i<n; i++){
		tree[i][0]=i;
		tree[i][1]=1;
		tree[i][2]=1;
	}
	for(i=0; i<m; i++){
		scanf("%d%d", &a[i][0], &a[i][1]);
		a[i][0]--; a[i][1]--;
		unon(a[i][0], a[i][1]);
	}
	for(i=0; i<n; i++){
		ans[i]=tree[root(i)][2]-1;
	}
	for(i=0; i<m; i++){
		if(tree[a[i][0]][0]==tree[a[i][1]][0]){
			ans[a[i][0]]--;
			ans[a[i][1]]--;
		}
	}
	for(i=0; i<k; i++){
		int b[2];
		scanf("%d%d", &b[0], &b[1]);
		b[0]--; b[1]--;
		if(tree[b[0]][0]==tree[b[1]][0]){
			ans[b[0]]--;
			ans[b[1]]--;
		}
	}
	for(i=0; i<n; i++){
		printf("%d ", ans[i]);
	} putchar(10);
}