#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Up, typename... Down>
void dbg_out(Up U, Down... D) {
  cerr << ' ' << U;
  dbg_out(D...);
}
long long fre[100001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j, t, k, n, x, y, z, num, c, cnt = 0;
  set<int> st;
  map<int, int> m;
  vector<vector<string>> all;
  vector<long long> v, v1;
  string s1, s2, s3;
  pair<int, int> p[100000];
  deque<char> deq;
  vector<double> d;
  cin >> t;
  while (t--) {
    cin >> n;
    string s;
    vector<string> ar;
    for (i = 0; i < 2; i++) cin >> s, ar.push_back(s);
    bool flag = true;
    for (j = 0; j < n; j++) {
      if (ar[0][j + 1] == '1' && ar[0 + 1][j + 1] == '1') {
        flag = false;
        break;
      } else
        continue;
    }
    if (flag)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

### 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 A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Up, typename... Down>
void dbg_out(Up U, Down... D) {
  cerr << ' ' << U;
  dbg_out(D...);
}
long long fre[100001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long i, j, t, k, n, x, y, z, num, c, cnt = 0;
  set<int> st;
  map<int, int> m;
  vector<vector<string>> all;
  vector<long long> v, v1;
  string s1, s2, s3;
  pair<int, int> p[100000];
  deque<char> deq;
  vector<double> d;
  cin >> t;
  while (t--) {
    cin >> n;
    string s;
    vector<string> ar;
    for (i = 0; i < 2; i++) cin >> s, ar.push_back(s);
    bool flag = true;
    for (j = 0; j < n; j++) {
      if (ar[0][j + 1] == '1' && ar[0 + 1][j + 1] == '1') {
        flag = false;
        break;
      } else
        continue;
    }
    if (flag)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

```