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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

type UnionFind=record n:Longint;pr:Array of Longint;end;
procedure init(var uf:UnionFind;N:Longint);
var i:Longint;
begin
	uf.n:=N;
	setlength(uf.pr,N);
	for i:=1 to N do uf.pr[i]:=-1;
end;
function find(var uf:UnionFind;a:Longint):Longint;
begin
	if uf.pr[a]<0 then find:=a else begin
		uf.pr[a]:=find(uf,uf.pr[a]);
		find:=uf.pr[a];
	end;
end;
function same(var uf:UnionFind;a,b:Longint):Boolean;
begin
	same:=find(uf,a)=find(uf,b);
end;
function unite(var uf:UnionFind;a,b:Longint):Boolean;
begin
	a:=find(uf,a);
	b:=find(uf,b);
	if a=b then unite:=false else begin
		if uf.pr[a]>uf.pr[b] then begin
			inc(uf.pr[b],uf.pr[a]);
			uf.pr[a]:=b;
		end else begin
			inc(uf.pr[a],uf.pr[b]);
			uf.pr[b]:=a;
		end;
		unite:=true;
	end;
end;
function size(var uf:UnionFind;a:Longint):Longint;
begin
	size:=-uf.pr[find(uf,a)];
end;
var
	U,V,suc,beg:Array[1..400000]of Longint;
	N,M,K,i,id,a,b,ec,ans:Longint;
	uf:UnionFind;
procedure add_edge(a,b:Longint);
begin
	inc(ec);
	U[ec]:=a;
	V[ec]:=b;
	suc[ec]:=beg[a];
	beg[a]:=ec;
end;
begin
	read(N,M,K);
	init(uf,N);
	for i:=1 to M do begin
		read(a,b);
		add_edge(a,b);
		add_edge(b,a);
		unite(uf,a,b);
	end;
	for i:=1 to K do begin
		read(a,b);
		add_edge(a,b);
		add_edge(b,a);
	end;
	for i:=1 to N do begin
		ans:=size(uf,i)-1;
		id:=beg[i];
		while id>0 do begin
			if same(uf,i,V[id]) then dec(ans);
			id:=suc[id];
		end;
		writeln(ans);
	end;
end.