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;
  const long long mod = 1e9 + 7;
  const long long MM = 998244353;
  const long long inf = LLONG_MAX;
  const long long dif = LLONG_MIN;
  static mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename A>
  ostream& operator<<(ostream& cout, vector<A> const& v);
  template <typename A, typename B>
  ostream& operator<<(ostream& cout, pair<A, B> const& p) {
    return cout << p.first << " " << p.second;
  }
  template <typename A>
  ostream& operator<<(ostream& cout, vector<A> const& v) {
    for (long long i = 0; i < (long long)((v).size()); ++i) {
      cout << v[i] << " ";
    }
    return cout << "\n";
  }
  template <typename A, typename B>
  istream& operator>>(istream& cin, pair<A, B>& p) {
    cin >> p.first;
    return cin >> p.second;
  }
  template <typename A>
  istream& operator>>(istream& cin, vector<A>& v) {
    for (auto& x : v) cin >> x;
    return cin;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p);
  string to_string(const string& s) { return '"' + s + '"'; }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  string to_string(vector<bool> v) {
    bool first = true;
    string res = "{";
    for (long long i = 0; i < static_cast<long long>(v.size()); ++i) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(v[i]);
    }
    res += "}";
    return res;
  }
  template <size_t N>
  string to_string(bitset<N> v) {
    string res = "";
    for (size_t i = 0; i < N; ++i) {
      res += static_cast<char>('0' + v[i]);
    }
    return res;
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  void dbg_out() { cout << "\n"; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cout << " " << to_string(H);
    dbg_out(T...);
  }
  void solve() {
    long long n;
    string a, b;
    cin >> n >> a >> b;
    bool safe = 1;
    for (long long i = 0; i < n; ++i) {
      if (a[i] == '1' and a[i] == b[i]) {
        safe = 0;
        break;
      }
    }
    cout << (safe ? "YES\n" : "NO\n");
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    long long tc = 1;
    cin >> tc;
    while (tc--) solve();
  }
  