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 9.2.1) (judge verdict: Accepted)

#include <algorithm>
#include <cmath>
#include <vector>
#include <functional>
#include <cstdlib>
#include <map>
#include <set>
#include <iostream>
#include <string>
#include <ctype.h>
#include <climits>
#include <queue>
using namespace std;
typedef long long ll;
#define REP(i, n) for(ll i = 0; i < (ll)(n); ++i)
template<class T> inline bool chmax(T& a, T b) { if(a < b){ a=b; return 1; } return 0;}
template<class T> inline bool chmin(T& a, T b) { if(a > b){ a=b; return 1; } return 0;}

const int MAXN=100000;
int par[MAXN];
int rnk[MAXN];

void init(int n){
	REP(i,n){
		par[i] = i;
		rnk[i] = 0;
	}
}
int find(int x){
	if(par[x]==x){
		return x;
	}else{
		return par[x]=find(par[x]);
	}
}
void unite(int x, int y){
	x = find(x);
	y = find(y);
	if(x==y) return;
	if(rnk[x] < rnk[y]){
		par[x] = y;
	}else{
		par[y] = x;
		if(rnk[x]==rnk[y]) rnk[x]++;
	}
}
bool same(int x,int y){
	return find(x)==find(y);
}

int main(){
	int n,m,k;
	cin >> n >> m >> k;
	vector<pair<int, int>> fri(m),blo(k);
	REP(i,m){
		cin >> fri[i].first >> fri[i].second;
		fri[i].first--; fri[i].second--;
	}
	REP(i,k){
		cin >> blo[i].first >> blo[i].second;
		blo[i].first--; blo[i].second--;
	}
	
	init(n);
	vector<int> ans(n,0);
	map<int,int> member;
	REP(i,m){
		int a = fri[i].first;
		int b = fri[i].second;
		ans[a]++; ans[b]++;
		unite(a,b);
	}
	REP(i,n){
		member[find(i)]++;
	}
	REP(i,k){
		int a = blo[i].first;
		int b = blo[i].second;
		if(same(a,b)){
			ans[a]++; ans[b]++;
		}
	}
	
	REP(i,n){
		if(i!=0) cout << " ";
		int a = member[find(i)];
		cout << a - ans[i] -1;
	}
	cout << endl;
	
	
	return 0;
}