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 <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <numeric>
#include <random>
#include <vector>
#include <array>
#include <bitset>
#include <queue>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n==0) ? 1 : 10*TEN(n-1); }
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;


bool solve() {
    int n;
    cin >> n;

    VV<int> ls;
    map<V<int>, int> mp;
    for (int i = 0; i < n; i++) {
        int w;
        cin >> w;
        V<int> v(3);
        iota(begin(v), end(v), 0);
        for (int j = 0; j < w; j++) {
            int a;
            cin >> a;
            swap(v[2*a], v[1]);
        }
        ls.push_back(v);
        mp[v]++;
        if (v == V<int>{0, 1, 2}) return true;
        if (v == V<int>{1, 0, 2} && mp[v] >= 2) return true;
        if (v == V<int>{0, 2, 1} && mp[v] >= 2) return true;
        if (v == V<int>{2, 1, 0} && mp[v] >= 2) return true;
        if (mp[v] >= 3) return true;
    }
    sort(begin(ls), end(ls));
    do {
        V<int> v(3), v2(3);
        iota(begin(v), end(v), 0);
        for (auto &w: ls) {
            for (int i = 0; i < 3; i++) {
                v2[w[i]] = v[i];
            }
            for (int i = 0; i < 3; i++) {
                v[i] = v2[i];
            }
            if (v == V<int>{0, 1, 2}) return true;
        }

    } while (next_permutation(begin(ls), end(ls)));
    return false;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(20);

    if (solve()) {
        cout << "yes" << endl;
    } else {
        cout << "no" << endl;
    }

    return 0;

}