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;
  template <class T>
  ostream &operator<<(ostream &os, const vector<T> &p) {
    os << "[";
    for (auto &it : p) os << it << " ";
    return os << "]";
  }
  template <class S, class T>
  ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << "(" << p.first << "," << p.second << ")";
  }
  template <class T>
  inline T Bit(T x, long long i) {
    return (x >> i) & 1;
  }
  template <class T>
  bool umin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool umax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  const long long N = 2e5 + 3;
  void solve(long long tc) {
    long long n;
    cin >> n;
    string s[2];
    for (long long i = 0; i < 2; i++) cin >> s[i];
    long long a[2][n];
    memset(a, 0, sizeof(a));
    a[0][0] = 1;
    if (s[1][0] == '0') a[1][0] = 1;
    for (long long i = 1; i < n; i++) {
      if (a[0][i - 1] || a[1][i - 1]) {
        if (s[0][i] == '0') a[0][i] = 1;
        if (s[1][i] == '0') a[1][i] = 1;
      }
    }
    if (a[1][n - 1]) {
      cout << "YES" << '\n';
      return;
    };
    {
      cout << "NO" << '\n';
      return;
    };
  }
  signed main() {
    ios_base ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    for (long long i = 1; i <= t; i++) solve(i);
    return 0;
  }
  