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 Solution {
  
      /* -------------------------------------- Main Start ----------------------------------- */
      
      public static void main(String[] args) throws IOException {
          FastReader fs = new FastReader();
          StringBuilder sb = new StringBuilder();
          int t = Integer.parseInt(fs.nextLine());
          while (t-- > 0) {
              int n = fs.nextInt();
              char[][] pos = new char[2][n];
              pos[0] = fs.nextLine().toCharArray();
              pos[1] = fs.nextLine().toCharArray();
              if(pos[0][0] == '1' || pos[1][n - 1] == '1'){
                  System.out.println("NO");
                  continue;
              }
              boolean isPoss = true;
              for(int i = 1; i < n - 1; i++){
                  if(pos[0][i] == '1' && pos[1][i] == '1'){
                      isPoss = false;
                      break;
                  }
              }
              if(isPoss) System.out.println("YES");
              else System.out.println("NO");
          }
      }
    
      /* ---------------------------------------------------------------------------------------------------------------------------------------------------------------------- 
  
        StringBuilder sb = new StringBuilder();
        sb.append(x + "\n");
      
        Collections.sort(arr, (a, b) -> Integer.compare(a[0], b[0]))
        arr.toArray(new int[arr.size()][])
      
       */
      private static final int MOD_1 = 1000000000 + 7;
  
      static class FastReader {
  
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() throws IOException {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() throws IOException {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() throws IOException {
              return Integer.parseInt(next());
          }
  
          long nextLong() throws IOException {
              return Long.parseLong(next());
          }
  
          double nextDouble() throws IOException {
              return Double.parseDouble(next());
          }
  
          String nextLine() throws IOException {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          /* --------------------------------------------------------------------------------------------------------------------------------------------------------------------- */
          void printArray(int[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void printArray(String[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void printArray(long[] arr) {
              System.out.println(Arrays.toString(arr));
          }
  
          void print(int data) {
              System.out.println(data);
          }
  
          void print(String data) {
              System.out.println(data);
          }
  
          void print(long data) {
              System.out.println(data);
          }
  
          int[] II(int n) throws IOException {
              int[] d = new int[n];
              String[] arr = nextLine().split(" ");
              for (int i = 0; i < n; i++) {
                  d[i] = Integer.parseInt(arr[i]);
              }
  
              return d;
          }
  
          String[] IS(int n) throws IOException {
              return nextLine().split(" ");
          }
  
          long[] IL(int n) throws IOException {
              long[] d = new long[n];
              String[] arr = nextLine().split(" ");
              for (int i = 0; i < n; i++) {
                  d[i] = Long.parseLong(arr[i]);
              }
              return d;
          }
  
          public long gcd(long a, long b) {
              if (a == 0) {
                  return b;
              }
              return gcd(b % a, a);
          }
  
          public long power(long x, long y, long p) {
              long res = 1;
              x = x % p;
              if (x == 0) {
                  return 0;
              }
  
              while (y > 0) {
                  if ((y & 1) == 1) {
                      res = (res * x) % p;
                  }
                  y = y >> 1;
                  x = (x * x) % p;
              }
              return res;
          }
  
          void sieveOfEratosthenes(boolean prime[], int size) {
              Arrays.fill(prime, true);
              prime[0] = prime[1] = false;
              prime[2] = true;
  
              for (int p = 2; p * p < size; p++) {
                  if (prime[p] == true) {
                      for (int i = p * p; i < size; i += p) {
                          prime[i] = false;
                      }
                  }
              }
          }
  
          public long fact(long n) {
              long ans = 1;
              for (int i = 2; i <= n; i++) {
                  ans = (ans * i) % MOD_1;
              }
              return ans;
          }
  
          public long lcm(long a, long b) {
              return (a * b) / gcd(a, b);
          }
  
          public int log2(int N) {
              int result = (int) (Math.log(N) / Math.log(2));
              return result;
          }
  
          // find first element in the arr that is equal to the x. 
          // If x is not present in arr then it will return first element that is greater than x
          // It return arr.size() if all element are smaller than x
          public int lowerBound(int[] arr, int x) {
              int low = 0, high = arr.length, mid = -1;
  
              while (low < high) {
                  mid = (low + high) / 2;
  
                  if (arr[mid] >= x) {
                      high = mid;
                  } else {
                      low = mid + 1;
                  }
              }
  
              return low;
          }
  
          // find first element in the arr that is greater than x. 
          // It return arr.size() if all element are smaller than x
          public int upperBound(int[] arr, int x) {
              int low = 0, high = arr.length, mid = -1;
  
              while (low < high) {
                  mid = (low + high) / 2;
  
                  if (arr[mid] > x) {
                      high = mid;
                  } else {
                      low = mid + 1;
                  }
              }
  
              return low;
          }
  
      }
  
  }
  