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:
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

map<VI, int> mp;
signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  REP(i, n) {
    int w;
    cin >> w;
    VI vec = {0, 1, 2};
    REP(j, w) {
      int a; cin >> a;
      if(a == 0) {
        swap(vec[0], vec[1]);
      } else{
        swap(vec[1], vec[2]);
      }
    }
    mp[vec]++;
  }

  if(n >= 6) {
    cout << "yes" << endl;
    return 0;
  }

  VVI vv;
  for(auto i: mp) {
    REP(j, i.second) {
      vv.PB(i.first);
    }
  }

  sort(ALL(vv));
  FOR(i, 1, 1<<vv.size()) {
    VVI v;
    REP(j, vv.size()) {
      if(i&1<<j) v.PB(vv[j]);
    }
    // sort(ALL(v));
    // cout << v << endl;
    do {
      VI v2 = {0, 1, 2};
      for(auto &i: v) {
        v2[0] = i[v2[0]];
        v2[1] = i[v2[1]];
        v2[2] = i[v2[2]];
      }
      if(v2 == VI{0, 1, 2}) {
        cout << "yes" << endl;
        return 0;
      }
    } while(next_permutation(ALL(v)));
  }

  cout << "no" << endl;

  return 0;
}