#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
using namespace std;
bool ans;
void dfs(vector<string> &grid, int i, int j, vector<vector<int>> &vis) {
  if (i < 0 || i > 1 || j >= grid[0].size() || j < 0 || grid[i][j] == '1' ||
      vis[i][j] == 1 || ans == true)
    return;
  vis[i][j] = 1;
  if (i == 1 && j == grid[0].size() - 1) {
    ans = true;
    return;
  }
  int dx[8] = {1, -1, 0, 0, 1, -1, -1, 1};
  int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
  for (int ii = 0; ii < 8; ii++) {
    int xx = i + dx[ii], yy = j + dy[ii];
    if (xx < 0 || xx > 1 || yy < 0 || yy >= grid[0].size() ||
        grid[xx][yy] == '1' || vis[xx][yy] == 1)
      continue;
    dfs(grid, xx, yy, vis);
  }
  return;
}
void solve() {
  long long n;
  cin >> n;
  vector<vector<int>> vis(2, vector<int>(n, 0));
  vector<string> grid;
  string s1, s2;
  cin >> s1 >> s2;
  grid.push_back(s1);
  grid.push_back(s2);
  ans = false;
  dfs(grid, 0, 0, vis);
  cout << (ans == true ? "YES" : "NO") << "\n";
}
int main() {
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
using namespace std;
bool ans;
void dfs(vector<string> &grid, int i, int j, vector<vector<int>> &vis) {
  if (i < 0 || i > 1 || j >= grid[0].size() || j < 0 || grid[i][j] == '1' ||
      vis[i][j] == 1 || ans == true)
    return;
  vis[i][j] = 1;
  if (i == 1 && j == grid[0].size() - 1) {
    ans = true;
    return;
  }
  int dx[8] = {1, -1, 0, 0, 1, -1, -1, 1};
  int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
  for (int ii = 0; ii < 8; ii++) {
    int xx = i + dx[ii], yy = j + dy[ii];
    if (xx < 0 || xx > 1 || yy < 0 || yy >= grid[0].size() ||
        grid[xx][yy] == '1' || vis[xx][yy] == 1)
      continue;
    dfs(grid, xx, yy, vis);
  }
  return;
}
void solve() {
  long long n;
  cin >> n;
  vector<vector<int>> vis(2, vector<int>(n, 0));
  vector<string> grid;
  string s1, s2;
  cin >> s1 >> s2;
  grid.push_back(s1);
  grid.push_back(s2);
  ans = false;
  dfs(grid, 0, 0, vis);
  cout << (ans == true ? "YES" : "NO") << "\n";
}
int main() {
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```