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;
  void call() {
    int n;
    cin >> n;
    vector<vector<int>> arr(2, vector<int>(n));
    vector<vector<bool>> visited(2, vector<bool>(n, false));
    string s1;
    cin >> s1;
    string s2;
    cin >> s2;
    for (int j = 0; j < n; ++j) {
      arr[0][j] = s1[j] - '0';
    }
    for (int j = 0; j < n; ++j) {
      arr[1][j] = s2[j] - '0';
    }
    queue<pair<int, int>> q;
    q.push({0, 0});
    visited[0][0] = 1;
    while (q.empty() == false) {
      int k = q.size();
      while (k--) {
        pair<int, int> curr = q.front();
        int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
        int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
        int x = curr.first;
        int y = curr.second;
        q.pop();
        for (int i = 0; i < 8; ++i) {
          if (x + dx[i] >= 0 and x + dx[i] < 2 and y + dy[i] >= 0 and
              y + dy[i] < n and arr[x + dx[i]][y + dy[i]] == 0 and
              visited[x + dx[i]][y + dy[i]] == 0) {
            visited[x + dx[i]][y + dy[i]] = 1;
            q.push({x + dx[i], y + dy[i]});
          }
        }
      }
    }
    if (visited[1][n - 1]) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) call();
  }
  