思路参考上面 bilibili 的视频链接。

如果 $x | y = x$，那么 $y$ 这个数字是可有可无的，如果能够在数组中找到任何一个这样的 $y$，那么就存在解：子序列 $a$ 是所有的元素，子序列 $b$ 是除了 $y$ 之外的所有元素。$y$ 的特征是它所有的位在其它元素中存在至少一次，也就是 $y$ 的二进制位在所有元素中出现了至少两次。如果 $y_1$ 有一个位是独一无二的，那么一定不满足 $x | y_1 = x$，所以只需要找到一个所有二进制位在所有元素中出现至少两次的元素。

// Date: Sun Dec 10 14:25:09 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 100010;
int t, n;

int main(void) {
#ifdef _DEBUG
  freopen("1775b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    vector<VI> a(n);

    map<int, int> m;

    For(i, 0, n) {
      int len;
      cin >> len;

      For1(j, 1, len) {
        int x;
        cin >> x;

        m[x]++;
        a[i].pb(x);
      }
    }

    bool flag = false;

    For(i, 0, n) {
      bool mark = true;
      for (auto x : a[i]) {
        if (m[x] < 2) {
          mark = false;
          break;
        }
      }
      if (mark) {
        flag = true;
        break;
      }
    }

    cout << (flag ? "YES\n" : "NO\n");
  }

  return 0;
}