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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <array>
#include <bitset>
#include <assert.h>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

void calc(VI &x, VI y){
    REP(i,3) x[i] = y[x[i]];
}

int main() {
    int n;
    cin >> n;

    VI good(3);
    REP(i,3) good[i] = i;

    VVI p(n, VI(3));
    REP(i,n){
        int w = in();
        VI tmp(3);
        REP(j,3) tmp[j] = j;

        REP(j,w){
            int x = in();
            swap(tmp[x], tmp[x+1]);
        }

        if (tmp == good){
            cout << "yes" << endl;
            return 0;
        }

        REP(j,3){
            p[i][tmp[j]] = j;
        }
    }

    REP(i,n) REP(j,i){
        VI tmp(3);
        REP(k,3) tmp[k] = k;

        calc(tmp, p[i]);
        calc(tmp, p[j]);

        if (tmp == good){
            cout << "yes" << endl;
            return 0;
        }
    }

    // REP(i,n){
    //     REP(j,3) cout << p[i][j] << " ";
    //     cout << endl;
    // }

    REP(i,n) REP(j,n) REP(l,n){
        if (i == j || j == l || l == i) continue;
        VI tmp(3);
        REP(k,3) tmp[k] = k;

        calc(tmp, p[i]);
        calc(tmp, p[j]);
        calc(tmp, p[l]);


        if (tmp == good){
            cout << "yes" << endl;
            return 0;
        }
    }

    cout << "no" << endl;

    return 0;
}