Problem:
Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1

Solution in Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	N,A,x,y,i,j,out,ans,nowt,nowf,ccc:Longint;
	F,T,gF,gT:Array[1..15]of Longint;
procedure dfs(id,nt,nf:Longint);
var
cnt,i:Longint;
begin
	if id=N+1 then begin
		cnt:=0;
		for i:=1 to N do if((nt>>(i-1))and 1)=1 then inc(cnt);
		if ans<cnt then ans:=cnt;
	end else begin
		if((((nt or nf)>>(id-1))and 1)=0)and(((nf or gF[id])and(nt or gT[id]))=0)then dfs(id+1,nt or gT[id],nf or gF[id]);
		dfs(id+1,nt,nf);
	end;
end;
begin
	read(N);
	for i:=1 to N do begin
		read(A);
		for j:=1 to A do begin
			read(x,y);
			dec(x);
			if y=0 then F[i]:=F[i] or (1<<x)else T[i]:=T[i] or (1<<x);
		end;
	end;
	for i:=1 to N do begin
		nowt:=1<<(i-1);
		nowf:=0;
		for ccc:=1 to N do begin
			for j:=1 to N do begin
				if ((nowt>>(j-1))and 1)=1 then begin
					nowt:=nowt or T[j];
					nowf:=nowf or F[j];
				end;
			end;
		end;
		gF[i]:=nowf;
		gT[i]:=nowt;
	end;
	dfs(1,0,0);
	writeln(ans);
end.