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<bits/stdc++.h>
using namespace std;

long long int N, ami[6], a, w, p[3];

int main(){
    scanf("%lld", &N);
    for(int i = 0; i < N; ++i){
        scanf("%lld", &w);
        *p = 0;
        p[1] = 1;
        p[2] = 2;
        for(int j = 0; j < w; ++j){
            scanf("%lld", &a);
            swap(p[1], p[a * 2]);
        }
        w = *p * 2 + (p[1] > p[2]);
        //cout << w << endl;
        ++ami[w];
        switch(w){
            case 0:
                return 0 & puts("yes");
                break;
            case 1:
            case 2:
            case 5:
                if(ami[w] > 1)return 0 & puts("yes");
                break;
            default:
                if(ami[w] > 2)return 0 & puts("yes");
        }
    }
    if(ami[3] && ami[4])return 0 & puts("yes");
    if(ami[3] || ami[4]){
        if(ami[1] + ami[2] + ami[5] > 1)return 0 & puts("yes");
    }
    puts("no");
    return 0;
}