description: n students attended the first meeting of the Berland SU programming course (n is even). All students will be divided into two groups. Each group will be attending exactly one lesson each week during one of the five working days (Monday, Tuesday, Wednesday, Thursday and Friday), and the days chosen for the groups must be different. Furthermore, both groups should contain the same number of students.
  
  Each student has filled a survey in which they told which days of the week are convenient for them to attend a lesson, and which are not. 
  
  Your task is to determine if it is possible to choose two different week days to schedule the lessons for the group (the first group will attend the lesson on the first chosen day, the second group will attend the lesson on the second chosen day), and divide the students into two groups, so the groups have equal sizes, and for each student, the chosen lesson day for their group is convenient.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  Then the descriptions of t testcases follow.
  
  The first line of each testcase contains one integer n (2 ≤ n ≤ 1 000) — the number of students.
  
  The i-th of the next n lines contains 5 integers, each of them is 0 or 1. If the j-th integer is 1, then the i-th student can attend the lessons on the j-th day of the week. If the j-th integer is 0, then the i-th student cannot attend the lessons on the j-th day of the week. 
  
  Additional constraints on the input: for each student, at least one of the days of the week is convenient, the total number of students over all testcases doesn't exceed 10^5.
  
  Output
  
  For each testcase print an answer. If it's possible to divide the students into two groups of equal sizes and choose different days for the groups so each student can attend the lesson in the chosen day of their group, print "YES" (without quotes). Otherwise, print "NO" (without quotes). 
  
  Example
  
  Input
  
  
  2
  4
  1 0 0 1 0
  0 1 0 0 1
  0 0 0 1 0
  0 1 0 1 0
  2
  0 0 0 1 0
  0 0 0 1 0
  
  
  Output
  
  
  YES
  NO
  
  Note
  
  In the first testcase, there is a way to meet all the constraints. For example, the first group can consist of the first and the third students, they will attend the lessons on Thursday (the fourth day); the second group can consist of the second and the fourth students, and they will attend the lessons on Tuesday (the second day).
  
  In the second testcase, it is impossible to divide the students into groups so they attend the lessons on different days.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.File;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  public class Main {
   
  	
  	/* 10^(7) = 1s.
  	 * ceilVal = (a+b-1) / b */
  	
  	static final int mod = 1000000007;
  	static final long temp = 998244353;
  	static final long MOD = 1000000007;
  	static final long M = (long)1e9+7;
   
  	static class Pair implements Comparable<Pair> 
  	{
  		int first, second;
  		
  		public Pair(int first, int second) 
  		{
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair ob) 
  		{
  			return (int)(first - ob.first);
  		}
  	}
  	
  	
  	static class Tuple implements Comparable<Tuple>
  	{
  		int first, second,third;
  		public Tuple(int first, int second, int third)
  		{
  			this.first = first;
  			this.second = second;
  			this.third = third;
  		}
  		public int compareTo(Tuple o)
  		{
  			return (int)(o.third - this.third);
  		}
  	}
  	
  	public static class DSU
  	{
  		int[] parent;
  		int[] rank; //Size of the trees is used as the rank
  		public DSU(int n)
  		{
  			parent = new int[n];
  			rank = new int[n];
  			Arrays.fill(parent, -1);
  			Arrays.fill(rank, 1);
  		}
  		
  		public int find(int i) //finding through path compression
  		{
  			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
  		}
  		
  		public boolean union(int a, int b) //Union Find by Rank 
  		{
  			a = find(a);
  			b = find(b);
  			
  			if(a == b) return false; //if they are already connected we exit by returning false.
  			
  			// if a's parent is less than b's parent
  			if(rank[a] < rank[b])
  			{
  				//then move a under b
  				parent[a] = b;
  			}
  			//else if rank of j's parent is less than i's parent
  			else if(rank[a] > rank[b])
  			{
  				//then move b under a
  				parent[b] = a;
  			}
  			//if both have the same rank.
  			else
  			{
  				//move a under b (it doesnt matter if its the other way around.
  				parent[b] = a;
  				rank[a] = 1 + rank[a];
  			}
  			return true;
  		}	
  	}
  
  	static class Reader 
  	{
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) throws IOException {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		
  		long[] longReadArray(int n) throws IOException {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  	
  	public static int gcd(int a, int b)
  	{
  		if(b == 0)
  		 return a;
  		
  		else
  		return gcd(b,a%b);
  	} 
   
  		public static long lcm(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  	public static long LongGCD(long a, long b)
  	{
  		if(b == 0)
  			 return a;
  			
  			else
  			return LongGCD(b,a%b);
  	}
  	
  		public static long LongLCM(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  		//Count the number of coprime's upto N
  		public static long phi(long n)  //euler totient/phi function
  	    { 
  	       long ans = n;
  //	       for(long i = 2;i*i<=n;i++)
  //	       {
  //	    	   if(n%i == 0)
  //	    	   {
  //	    		   while(n%i == 0) n/=i;
  //	    		   ans -= (ans/i);
  //	    	   }
  //	       }
  //	       
  //	       if(n > 1)
  //	       {
  //	    	   ans -= (ans/n);
  //	       }
  	       
  	       for(long i = 2;i<=n;i++)
  	       {
  	    	   if(isPrime(i))
  	    	   {
  	    		   ans -= (ans/i);
  	    	   }
  	       }
  	       
  	       return ans;
  	       
  	    }	
  	
  	public static long fastPow(long x, long n) 
  	{
  		if(n == 0)
  			return 1;
  		else if(n%2 == 0)
  			return fastPow(x*x,n/2);
  		else
  			return x*fastPow(x*x,(n-1)/2);
  	}
  	
  	   public static long modPow(long x, long y, long p)
  	    {
  	        long res = 1;
  	        x = x % p;
  	 
  	        while (y > 0) {
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	        return res;
  	    }
  	   
  	   static long modInverse(long n, long p)
  	    {
  	        return modPow(n, p - 2, p);
  	    }
  	 
  	    // Returns nCr % p using Fermat's little theorem.
  	    
  	    public static long nCrModP(long n, long r,long p)
  	    {
  	    	if (n<r) 
  	              return 0;
  	          
  	        if (r == 0)
  	            return 1;
  	 
  	        long[] fac = new long[(int)(n) + 1];
  	        fac[0] = 1;
  	 
  	        for (int i = 1; i <= n; i++)
  	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
  	                % p * modInverse(fac[(int)(n - r)], p)
  	                % p)
  	            % p;
  	    }
  	    
  	    public static long fact(long n) 
  	    {
  	    	  long[] fac = new long[(int)(n) + 1];
  		        fac[0] = 1;
  		 
  		        for (long i = 1; i <= n; i++)
  		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
  		        
  		        return fac[(int)(n)];
  		}
  	    
  	    public static long nCr(long n, long k) 
  	    {
  			long ans = 1;
  			for(long i = 0;i<k;i++)
  			{
  				ans *= (n-i);
  				ans /= (i+1);
  			}
  			return ans;
  		}
   
  		//Modular Operations for Addition and Multiplication. 
  		   public static long perfomMod(long x)
  		   {
  		        return ((x%M + M)%M);
  		    }
  		   public static long addMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)+perfomMod(b));
  		    }
  		  
  		   public static long subMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)-perfomMod(b));
  		   }
  		   
  		   public static long mulMod(long  a, long b)
  		   {
  		        return perfomMod(perfomMod(a)*perfomMod(b));
  		   }
  		   
  		   public static boolean isPrime(long n) 
  			{
  				if(n == 1)
  				{
  					return false;
  				}
  				
  				//check only for sqrt of the number as the divisors 
  				//keep repeating so only half of them are required. So,sqrt.
  				for(int i = 2;i*i<=n;i++)
  				{
  					if(n%i == 0)
  					{
  						return false;
  					}
  				}
  				return true;
  			}
  			
  			public static List<Long> SieveList(int n)
  			{
  				boolean prime[] = new boolean[(int)(n+1)];
  				Arrays.fill(prime, true);
  				
  				List<Long> l = new ArrayList<>();
  				for (long p = 2; p*p<=n; p++) 
  				{ 
  					if (prime[(int)(p)] == true) 
  					{ 	
  					    for(long i = p*p; i<=n; i += p) 
  						{
  						    prime[(int)(i)] = false; 
  						}	
  					} 
  				} 
  		 
  				for (long p = 2; p<=n; p++) 
  				{
  				    if (prime[(int)(p)] == true)
  				    {
  				       l.add(p); 
  				    }
  				}
  				
  				return l;
  			}
  
  			public static long log2(long n)
  		    {
  				 long ans = (long)(log(n)/log(2));
  				 return ans;
  			}
  			   
  			public static boolean isPow2(long n)
  			{
  				 return (n  != 0 && ((n & (n-1))) == 0);
  			}	
  		   
  		   public static boolean isSq(int x)  
  		   {
  			    long s = (long)Math.round(Math.sqrt(x));
  			    return s*s==x;
  		   }
  		
  			   /*
  			    * 
  			    * >=         <=
  			   0 1 2 3 4 5 6 7  
  			   5 5 5 6 6 6 7 7
  			   lower_bound for 6 at index 3 (>=)
  			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
  			   */
  
  		   public static int LowerBound(long a[], long x) 
  		   {
  			      int l=-1,r=a.length;
  				  while(l+1<r) 
  				  {
  				    int m=(l+r)>>>1;
  				    if(a[m]>=x) r=m;
  				    else l=m;
  				  }
  				  return r;
  			}
  		   
  			public static int UpperBound(long a[], long x) 
  			{
  				int l=-1, r=a.length;
  			    while(l+1<r) 
  			    {
  			       int m=(l+r)>>>1;
  			       if(a[m]<=x) l=m;
  			       else r=m;
  			    }
  			    return l+1;
  			}
  			
  			public static void Sort(long[] a) 
  			{
  				List<Long> l = new ArrayList<>();
  				for (long i : a) l.add(i);
  				Collections.sort(l);
  				Collections.reverse(l);  //Use to Sort decreasingly 
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  			public static void ssort(char[] a)
  			{
  				List<Character> l = new ArrayList<>();
  				for (char i : a) l.add(i);
  				Collections.sort(l);
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  		
  	public static void main(String[] args) throws Exception 
  	{
  	
  		Reader sc = new Reader();
  		PrintWriter fout = new PrintWriter(System.out);
  
  		int t = sc.nextInt();
  		while(t-- > 0)
  		{
  			int n = sc.nextInt(), m = 5;
  			int[][] grid = new int[n][m];
  			for(int i = 0;i<n;i++)
  			{
  				for(int j = 0;j<m;j++) grid[i][j] = sc.nextInt();
  			}
  			
  			boolean ans = false;
  			
  			for(int i = 0;i<m;i++)
  			{
  				for(int j=i+1;j<m;j++)
  				{
  					int a = 0, b = 0, ab = 0;
  					for(int k = 0;k<n;k++)
  					{
  						if(grid[k][i] != 0)
  						{
  							a++;
  						}
  						if(grid[k][j] != 0)
  						{
  							b++;
  						}
  						if(grid[k][i] != 0 && grid[k][j] != 0)
  						{
  							ab++;
  						}
  					}
  					
  					if(a >= n/2 && b >= n/2 && a + b - ab == n) 
  					{
  						ans = true;
  						break;
  					}
  				}
  			}
  			
  			fout.println((ans == true) ? "YES" : "NO");
  		}
  		
  		fout.close();
  	}
  }