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: #include <bits/stdc++.h>
  using namespace std;
  int gcd(int a, int b) {
    int temp;
    while (b > 0) {
      temp = a % b;
      a = b;
      b = temp;
    }
    return a;
  }
  long long gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
  string to_binary(int x) {
    string s;
    while (x > 0) {
      s += (x % 2 ? '1' : '0');
      x /= 2;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  int binarySearch(int arr[], int l, int h, int x) {
    if (h >= l) {
      int mid = l + (h - l) / 2;
      if (arr[mid] == x) return mid;
      if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
      return binarySearch(arr, mid + 1, h, x);
    }
    return -1;
  }
  long long power(int A, long long B, int C) {
    if (A == 0) return 0;
    if (B == 0) return 1;
    long y;
    if (B % 2 == 0) {
      y = power(A, B / 2, C);
      y = (y * y) % C;
    } else {
      y = A % C;
      y = (y * power(A, B - 1, C) % C) % C;
    }
    return ((y) % C);
  }
  void yes() {
    cout << "YES"
         << "\n";
  }
  void no() {
    cout << "NO"
         << "\n";
  }
  long long f(long long a) {
    long long res[] = {a, 1, a + 1, 0};
    return res[a % 4];
  }
  long long getXor(long long a, long long b) { return f(b) ^ f(a - 1); }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      string s, s1;
      cin >> s >> s1;
      bool flag = true;
      if (s1[n - 1] == '1') flag = false;
      for (int i = 0; i < n; i++) {
        if (s[i] == '1' && s1[i] == '1') flag = false;
      }
      if (flag)
        yes();
      else
        no();
    }
    return 0;
    ;
  }
  