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.
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
/*
-> Give your 100%, that's it!
-> Rules To Solve Any Problem:
    1. Read the problem.
    2. Think About It.
    3. Solve it!
*/

public class Template {
    
	static int mod = 1000000007;

	public static void main(String[] args){
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int yo = sc.nextInt();
        while (yo-- > 0) {
            int n = sc.nextInt();
            int[][] arr = new int[n][5];
            for(int i = 0; i < n; i++){
                for(int j = 0; j < 5; j++){
                    arr[i][j] = sc.nextInt();
                }
            }
            
            boolean ok = false;
            ot: for(int i = 0; i < 5; i++){
                for(int j = 0; j < 5; j++){
                    if(i == j) continue;
                    Set<Integer> set1 = new HashSet<>();
                    Set<Integer> set2 = new HashSet<>();
                    Set<Integer> set3 = new HashSet<>();
                    for(int k = 0; k < n; k++){
                        if(arr[k][i] == 1 && arr[k][j] == 1){
                            set3.add(k);
                        }
                        else if(arr[k][i] == 1){
                            set1.add(k);
                        }
                        else if(arr[k][j] == 1){
                            set2.add(k);
                        }
                    }
                    int left = set1.size();
                    int right = set2.size();
                    int both = set3.size();
                    if(left + right + both == n && left <= n/2 && right <= n/2){
                        ok = true;
                        break ot;
                    }
                }
            }
            
            if(ok){
                out.println("YES");
            }
            else {
                out.println("NO");
            }
        }
        out.close();
	}
    
   
    
    
    /*
    Source: hu_tao
         Random stuff to try when stuck:
            -if it's 2C then it's dp
            -for combo/probability problems, expand the given form we're interested in
            -make everything the same then build an answer (constructive, make everything 0 then do something)
            -something appears in parts of 2 --> model as graph
            -assume a greedy then try to show why it works
            -find way to simplify into one variable if multiple exist
            -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
            -find lower and upper bounds on answer
            -figure out what ur trying to find and isolate it
            -see what observations you have and come up with more continuations
            -work backwards (in constructive, go from the goal to the start)
            -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
            -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
            -draw something
            -simulate a process
            -dont implement something unless if ur fairly confident its correct
            -after 3 bad submissions move on to next problem if applicable
            -do something instead of nothing and stay organized
            -write stuff down
         Random stuff to check when wa:
            -if code is way too long/cancer then reassess
            -switched N/M
            -int overflow
            -switched variables
            -wrong MOD
            -hardcoded edge case incorrectly
         Random stuff to check when tle:
            -continue instead of break
            -condition in for/while loop bad
         Random stuff to check when rte:
            -switched N/M
            -long to int/int overflow
            -division by 0
            -edge case for empty list/data structure/N=1
      */

	public static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	public static void sort(int[] arr) {
		ArrayList<Integer> ls = new ArrayList<Integer>();
		for (int x : arr)
			ls.add(x);
		Collections.sort(ls);
		for (int i = 0; i < arr.length; i++)
			arr[i] = ls.get(i);
	}

	public static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static boolean[] sieve(int N) {
		boolean[] sieve = new boolean[N + 1];
		for (int i = 2; i <= N; i++)
			sieve[i] = true;

		for (int i = 2; i <= N; i++) {
			if (sieve[i]) {
				for (int j = 2 * i; j <= N; j += i) {
					sieve[j] = false;
				}
			}
		}
		return sieve;
	}

	public static long power(long x, long y, long p) {
		long res = 1L;
		x = x % p;
		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y >>= 1;
			x = (x * x) % p;
		}
		return res;
	}

	public static void print(int[] arr, PrintWriter out) {
		//for debugging only
		for (int x : arr)
			out.print(x + " ");
		out.println();
	}

	static class FastScanner {
		private int BS = 1 << 16;
		private char NC = (char) 0;
		private byte[] buf = new byte[BS];
		private int bId = 0, size = 0;
		private char c = NC;
		private double cnt = 1;
		private BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) {
			try {
				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
			} catch (Exception e) {
				in = new BufferedInputStream(System.in, BS);
			}
		}

		private char getChar() {
			while (bId == size) {
				try {
					size = in.read(buf);
				} catch (Exception e) {
					return NC;
				}
				if (size == -1)
					return NC;
				bId = 0;
			}
			return (char) buf[bId++];
		}

		public int nextInt() {
			return (int) nextLong();
		}

		public int[] readInts(int N) {
			int[] res = new int[N];
			for (int i = 0; i < N; i++) {
				res[i] = (int) nextLong();
			}
			return res;
		}

		public long[] readLongs(int N) {
			long[] res = new long[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextLong();
			}
			return res;
		}

		public long nextLong() {
			cnt = 1;
			boolean neg = false;
			if (c == NC)
				c = getChar();
			for (; (c < '0' || c > '9'); c = getChar()) {
				if (c == '-')
					neg = true;
			}
			long res = 0;
			for (; c >= '0' && c <= '9'; c = getChar()) {
				res = (res << 3) + (res << 1) + c - '0';
				cnt *= 10;
			}
			return neg ? -res : res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c != '.' ? cur : cur + nextLong() / cnt;
		}

		public double[] readDoubles(int N) {
			double[] res = new double[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextDouble();
			}
			return res;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c > 32) {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c != '\n') {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if (c > 32)
				return true;
			while (true) {
				c = getChar();
				if (c == NC)
					return false;
				else if (c > 32)
					return true;
			}
		}
	}

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}