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<vector>
#include<algorithm>
#include<utility>
using namespace std;
int main () {
    int n;
    cin >> n;
    if (n > 5) {
        cout << "yes" << endl;
        return 0;
    }
    int rots[5][3];
    for(int i = 0; i < n; i++) {
        int m;
        cin >> m;
        for(int j = 0; j < 3; j++) rots[i][j] = j;
        for(int j = 0; j < m; j++) {
            int b;
            cin >> b;
            if (b == 0) {
                swap(rots[i][0], rots[i][1]);
            } else {
                swap(rots[i][1], rots[i][2]);
            }
        }
    }
    int par[n];
    for(int i = 0; i < n; i++) par[i] = i;
    int ans[3] = {0, 1, 2};
    do {
        for(int i = 0; i < 3; i++) ans[i] = i;
        for(int i = 0; i < n; i++) {
            int tmp[3];
            for(int j = 0; j < 3; j++) tmp[j] = ans[j];
            for(int j = 0; j < 3; j++) ans[j] = tmp[rots[par[i]][j]];
            if (ans[0] == 0 && ans[1] == 1 && ans[2] == 2) {
                cout << "yes" << endl;
                return 0;
            }
        }
    } while(next_permutation(par, par + n));
    cout << "no" << endl;
    return 0;
}