#include <bits/stdc++.h>
using namespace std;
template <typename T>
using vv = vector<vector<T>>;
void scar(long long *arr, long long n) {
  for (long long i = 0; i < n; i++) cin >> arr[i];
}
vector<pair<long long, long long>> px({{-1, 1}, {0, 1}, {1, 1}});
void q1() {
  long long n;
  string str[2];
  cin >> n;
  cin >> str[0] >> str[1];
  long long x = 0, y = 0;
  if (str[0][0] == '1') {
    cout << "NO"
         << "\n";
    return;
  }
  vector<vector<bool>> vis(2, vector<bool>(n, false));
  vis[0][0] = true;
  queue<pair<long long, long long>> q;
  q.push({x, y});
  while (!q.empty()) {
    pair<long long, long long> top = q.front();
    q.pop();
    if (top.first == 1 && top.second == (n - 1)) {
      cout << "YES"
           << "\n";
      return;
    }
    for (auto p : px) {
      if (p.first + top.first >= 0 && p.first + top.first < 2 &&
          p.second + top.second < n &&
          vis[p.first + top.first][p.second + top.second] == false &&
          str[p.first + top.first][p.second + top.second] == '0') {
        vis[p.first + top.first][p.second + top.second] = true;
        q.push({p.first + top.first, p.second + top.second});
      }
    }
  }
  cout << "NO"
       << "\n";
}
void q2() {}
void q3() {}
void q4() {}
void q5() {}
void solve() { q1(); }
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using vv = vector<vector<T>>;
void scar(long long *arr, long long n) {
  for (long long i = 0; i < n; i++) cin >> arr[i];
}
vector<pair<long long, long long>> px({{-1, 1}, {0, 1}, {1, 1}});
void q1() {
  long long n;
  string str[2];
  cin >> n;
  cin >> str[0] >> str[1];
  long long x = 0, y = 0;
  if (str[0][0] == '1') {
    cout << "NO"
         << "\n";
    return;
  }
  vector<vector<bool>> vis(2, vector<bool>(n, false));
  vis[0][0] = true;
  queue<pair<long long, long long>> q;
  q.push({x, y});
  while (!q.empty()) {
    pair<long long, long long> top = q.front();
    q.pop();
    if (top.first == 1 && top.second == (n - 1)) {
      cout << "YES"
           << "\n";
      return;
    }
    for (auto p : px) {
      if (p.first + top.first >= 0 && p.first + top.first < 2 &&
          p.second + top.second < n &&
          vis[p.first + top.first][p.second + top.second] == false &&
          str[p.first + top.first][p.second + top.second] == '0') {
        vis[p.first + top.first][p.second + top.second] = true;
        q.push({p.first + top.first, p.second + top.second});
      }
    }
  }
  cout << "NO"
       << "\n";
}
void q2() {}
void q3() {}
void q4() {}
void q5() {}
void solve() { q1(); }
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```