description: Monocarp is playing a computer game. Now he wants to complete the first level of this game.
  
  A level is a rectangular grid of 2 rows and n columns. Monocarp controls a character, which starts in cell (1, 1) — at the intersection of the 1-st row and the 1-st column.
  
  Monocarp's character can move from one cell to another in one step if the cells are adjacent by side and/or corner. Formally, it is possible to move from cell (x_1, y_1) to cell (x_2, y_2) in one step if |x_1 - x_2| ≤ 1 and |y_1 - y_2| ≤ 1. Obviously, it is prohibited to go outside the grid.
  
  There are traps in some cells. If Monocarp's character finds himself in such a cell, he dies, and the game ends.
  
  To complete a level, Monocarp's character should reach cell (2, n) — at the intersection of row 2 and column n.
  
  Help Monocarp determine if it is possible to complete the level.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow. Each test case consists of three lines.
  
  The first line contains a single integer n (3 ≤ n ≤ 100) — the number of columns.
  
  The next two lines describe the level. The i-th of these lines describes the i-th line of the level — the line consists of the characters '0' and '1'. The character '0' corresponds to a safe cell, the character '1' corresponds to a trap cell.
  
  Additional constraint on the input: cells (1, 1) and (2, n) are safe.
  
  Output
  
  For each test case, output YES if it is possible to complete the level, and NO otherwise.
  
  Example
  
  Input
  
  
  4
  3
  000
  000
  4
  0011
  1100
  4
  0111
  1110
  6
  010101
  101010
  
  
  Output
  
  
  YES
  YES
  NO
  YES
  
  Note
  
  Consider the example from the statement.
  
  In the first test case, one of the possible paths is (1, 1) → (2, 2) → (2, 3).
  
  In the second test case, one of the possible paths is (1, 1) → (1, 2) → (2, 3) → (2, 4).
  
  In the fourth test case, one of the possible paths is (1, 1) → (2, 2) → (1, 3) → (2, 4) → (1, 5) → (2, 6).
solution: 
  import java.io.*;
  import java.util.*;
  public class cp
  {
  	static class Reader {
  		final private int BUFFER_SIZE = 1 << 16;
  		private DataInputStream din;
  		private byte[] buffer;
  		private int bufferPointer, bytesRead;
  
  		public Reader()
  		{
  			din = new DataInputStream(System.in);
  			buffer = new byte[BUFFER_SIZE];
  			bufferPointer = bytesRead = 0;
  		}
  
  		public Reader(String file_name) throws IOException
  		{
  			din = new DataInputStream(
  					new FileInputStream(file_name));
  			buffer = new byte[BUFFER_SIZE];
  			bufferPointer = bytesRead = 0;
  		}
  
  		public String readLine() throws IOException
  		{
  			byte[] buf = new byte[64]; // line length
  			int cnt = 0, c;
  			while ((c = read()) != -1) {
  				if (c == '\n') {
  					if (cnt != 0) {
  						break;
  					}
  					else {
  						continue;
  					}
  				}
  				buf[cnt++] = (byte)c;
  			}
  			return new String(buf, 0, cnt);
  		}
  
  		public int nextInt() throws IOException
  		{
  			int ret = 0;
  			byte c = read();
  			while (c <= ' ') {
  				c = read();
  			}
  			boolean neg = (c == '-');
  			if (neg)
  				c = read();
  			do {
  				ret = ret * 10 + c - '0';
  			} while ((c = read()) >= '0' && c <= '9');
  
  			if (neg)
  				return -ret;
  			return ret;
  		}
  
  		public long nextLong() throws IOException
  		{
  			long ret = 0;
  			byte c = read();
  			while (c <= ' ')
  				c = read();
  			boolean neg = (c == '-');
  			if (neg)
  				c = read();
  			do {
  				ret = ret * 10 + c - '0';
  			} while ((c = read()) >= '0' && c <= '9');
  			if (neg)
  				return -ret;
  			return ret;
  		}
  
  		public double nextDouble() throws IOException
  		{
  			double ret = 0, div = 1;
  			byte c = read();
  			while (c <= ' ')
  				c = read();
  			boolean neg = (c == '-');
  			if (neg)
  				c = read();
  
  			do {
  				ret = ret * 10 + c - '0';
  			} while ((c = read()) >= '0' && c <= '9');
  
  			if (c == '.') {
  				while ((c = read()) >= '0' && c <= '9') {
  					ret += (c - '0') / (div *= 10);
  				}
  			}
  
  			if (neg)
  				return -ret;
  			return ret;
  		}
  
  		private void fillBuffer() throws IOException
  		{
  			bytesRead = din.read(buffer, bufferPointer = 0,
  					BUFFER_SIZE);
  			if (bytesRead == -1)
  				buffer[0] = -1;
  		}
  
  		private byte read() throws IOException
  		{
  			if (bufferPointer == bytesRead)
  				fillBuffer();
  			return buffer[bufferPointer++];
  		}
  
  		public void close() throws IOException
  		{
  			if (din == null)
  				return;
  			din.close();
  		}
  	}
  	
  	/* Iterative Function to calculate (x^y) in O(log y) */
  	  static long power(long x, long y, long p)
  	  {
  	    long res = 1; // Initialize result
  	 
  	    x = x % p; // Update x if it is more than or
  	    // equal to p
  	 
  	    if (x == 0)
  	      return 0; // In case x is divisible by p;
  	 
  	    while (y > 0)
  	    {
  	 
  	      // If y is odd, multiply x with result
  	      if ((y & 1) != 0)
  	        res = (res * x) % p;
  	 
  	      // y must be even now
  	      y = y >> 1; // y = y/2
  	      x = (x * x) % p;
  	    }
  	    return res;
  	  }
  	  
  	  static int power(int x,int y) {
  		  if(y == 0) {
  			  return 1;
  		  }
  		  if(y % 2 == 0) {
  			  int temp = power(x,y/2)%1000000007;
  			  return temp*temp%1000000007;
  			  
  		  }else {
  			  return ((x%1000000007)*power(x,y-1))%1000000007;
  		  }
  		  
  	  }
  	  
  
  
  	public static void main(String[] args) throws IOException
  	{	
  
  		try{
  			//Your Solve
  //			Reader s = new Reader();
  			Scanner s = new Scanner(System.in);
  			int t = s.nextInt();
  			for(int p = 0;p < t;p++) {
  				int n = s.nextInt();
  				String r1 = s.next();
  				String r2 = s.next();
  				int flag = 0;
  				for(int i = 0;i < n;i++) {
  					if((r1.charAt(i) == '0') || r2.charAt(i) == '0') {
  						
  					}else {
  						flag = 1;
  						break;
  					}
  				}
  				
  				if(flag == 1) {
  					System.out.println("NO");
  				}else {
  					System.out.println("YES");
  				}
  				
  				
  			}
  
  		}catch(Exception e){
  			return;
  		}
  	}
  }