Problem: Problem

There are $ N $ Amidakuji with 3 vertical lines.
No matter which line you start from, the Amidakuji that ends at the starting line is considered a good Amidakuji.
You can select one or more Amidakuji and connect them vertically in any order.
Output "yes" if you can make a good Amidakuji, otherwise output "no".

There are $ w_i $ horizontal lines in the $ i $ th Amidakuji.
$ a_ {i, j} $ indicates whether the $ j $ th horizontal bar from the top of Amidakuji $ i $ extends from the central vertical line to the left or right.
If $ a_ {i, j} $ is 0, it means that it extends to the left, and if it is 1, it means that it extends to the right.

Constraints

The input satisfies the following conditions.

* $ 1 \ le N \ le 50 $
* $ 0 \ le w_i \ le 100 $
* $ a_ {i, j} $ is 0 or 1

Input

The input is given in the following format.


$ N $
$ w_1 $ $ a_ {1,1} $ $ a_ {1,2} $ ... $ a_ {1, w_1} $
$ w_2 $ $ a_ {2,1} $ $ a_ {2,2} $ ... $ a_ {2, w_2} $
...
$ w_N $ $ a_ {N, 1} $ $ a_ {N, 2} $ ... $ a_ {N, w_N} $


All inputs are given as integers.
$ N $ is given on the first line.
$ W_i $ and $ w_i $ $ a_ {i, j} $ are given on the following $ N $ line, separated by blanks.

Output

If you can make a good Amidakuji, it outputs "yes", otherwise it outputs "no".

Examples

Input

3
2 0 0
1 1
5 1 0 0 1 0


Output

yes


Input

2
1 1
1 0


Output

no

Solution (in CPP):

Solution:
#include<iostream>
#include<set>
#include<vector>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
#define pb push_back
using namespace std;

int main(){
	int n;
	int check[6];
	rep(i,6)check[i]=0;
	cin>>n;
	rep(i,n){
		int w,a;
		cin>>w;
		int one=1,two=2,three=3;
		rep(j,w){
			cin>>a;
			if(a==0){
				int alt=one;
				one=two;
				two=alt;
			}else{
				int alt=two;
				two=three;
				three=alt;
			}
		}
		if(one==1){
			if(two==2)check[0]++;
			else check[2]++;
		}else if(one==2){
			if(two==1)check[1]++;
			else check[4]++;
		}else{
			if(two==1)check[5]++;
			else check[3]++;
		}
	}
	bool ans=false;
	if(check[0])ans=true;
	if(check[1]>1 || check[2]>1 || check[3]>1)ans=true;
	if(check[4] && check[5])ans=true;
	if(check[4]>2 || check[5]>2)ans=true;
	if((check[4] || check[5]) && ((check[1]&&check[2])||(check[2]&&check[3])||(check[3]&&check[1])))ans=true;
	cout<<(ans?"yes":"no")<<endl;
	return 0;
}